Every question on this topic with its answer, the working, and a full method: what the question is testing, the reason for each step, the mistakes that are easy to make and what they would give you, and the principle to take away.
1. INT-02
Find the area enclosed between the curve \(y = x^2 - 4x\) and the \(x\)-axis.
- A\(-\dfrac{32}{3}\)
- B\(\dfrac{8}{3}\)
- C\(8\)
- D\(\dfrac{16}{3}\)
- E\(\dfrac{64}{3}\)
- F\(\dfrac{32}{3}\)
Answer: F. Roots \(0\) and \(4\); the curve is below the axis between them. \(\int_0^4 (x^2 - 4x)\,dx = \frac{64}{3} - 32 = -\frac{32}{3}\), so the area is \(\frac{32}{3}\).
Full method — how and why
Area between a curve and the axis. The integral is negative here because the region is below the axis - the area is its magnitude.
- Find the limits: \(x^2 - 4x = x(x-4) = 0\) at \(x = 0\) and \(x = 4\).The enclosed region runs between consecutive roots.
- Note the curve is below the axis there: at \(x = 2\), \(y = -4\).Checking one interior value tells you the sign of the integral in advance.
- Integrate: \(\displaystyle\int_0^4 (x^2 - 4x)dx = \left[\frac{x^3}{3} - 2x^2\right]_0^4 = \frac{64}{3} - 32 = -\frac{32}{3}\).\(32 = \frac{96}{3}\), so \(\frac{64 - 96}{3} = -\frac{32}{3}\).
- Take the magnitude: the area is \(\dfrac{32}{3}\).Areas are positive; the negative sign only records that the region lies below the axis.
Where it goes wrong
- Giving \(-\frac{32}{3}\) as the area - it is offered precisely to catch this.
- Using limits \(0\) and \(2\) (the vertex) instead of the two roots.
Takeaway Integrate between the roots, then take the absolute value: a definite integral is a signed area.
2. INT-03
Given that \(\displaystyle\int_{1}^{5} f(x)\,dx = 7\) and \(\displaystyle\int_{3}^{5} f(x)\,dx = 2\), find \(\displaystyle\int_{1}^{3} \left(2f(x) - 1\right)dx\).
- A\(12\)
- B\(16\)
- C\(8\)
- D\(9\)
- E\(10\)
- F\(3\)
Answer: C. \(\int_1^3 f = 7 - 2 = 5\). So \(2 \times 5 - (3 - 1) = 8\).
Full method — how and why
Combining integrals over adjacent intervals, plus the linearity rules. No function is ever needed.
- Use \(\int_1^5 = \int_1^3 + \int_3^5\): so \(\int_1^3 f = 7 - 2 = 5\).Integrals over adjacent ranges add; this is the only way to get at the interval you need.
- Split the required integral: \(\int_1^3 (2f(x) - 1)dx = 2\int_1^3 f(x)dx - \int_1^3 1\,dx\).Constants come out; sums split - both are properties of the integral.
- Evaluate the second piece: \(\int_1^3 1\,dx = 3 - 1 = 2\).Integrating \(1\) over an interval gives its length.
- Combine: \(2(5) - 2 = 8\).Careful with the order of operations.
Where it goes wrong
- Forgetting the \(-1\) term and answering \(10\).
- Using \(\int_1^3 f = 7 + 2 = 9\) instead of subtracting.
Takeaway \(\int_a^c = \int_a^b + \int_b^c\), constants factor out, and \(\int_a^b 1\,dx = b - a\).
3. INT-04
A curve has \(\dfrac{dy}{dx} = 6x^2 - 4x + 1\) and passes through \((1, 3)\). Find its equation.
- A\(y = 12x - 4\)
- B\(y = 2x^3 - 4x^2 + x + 4\)
- C\(y = 2x^3 - 2x^2 + x + 2\)
- D\(y = 2x^3 - 2x^2 + x + 3\)
- E\(y = 2x^3 - 2x^2 + x\)
- F\(y = 2x^3 - 2x^2 + 2\)
Answer: C. \(y = 2x^3 - 2x^2 + x + c\); at \((1, 3)\): \(2 - 2 + 1 + c = 3 \Rightarrow c = 2\).
Full method — how and why
Reverse differentiation plus a boundary condition to pin down the constant.
- Integrate each term: \(\int(6x^2 - 4x + 1)dx = 2x^3 - 2x^2 + x + c\).Add one to each index and divide: \(\frac{6x^3}{3} = 2x^3\), \(\frac{-4x^2}{2} = -2x^2\).
- Do not forget the constant \(c\) - it is exactly what the given point determines.Every antiderivative differs by a constant, so one point selects the right curve.
- Substitute \((1,3)\): \(2 - 2 + 1 + c = 3\), so \(c = 2\).Evaluate the polynomial at \(x = 1\) and set it equal to \(3\).
- The curve is \(y = 2x^3 - 2x^2 + x + 2\).Check: differentiating gives back \(6x^2 - 4x + 1\) ✓.
Where it goes wrong
- Omitting \(c\) altogether and choosing \(y = 2x^3 - 2x^2 + x\).
- Arithmetic slip giving \(c = 3\) - an offered option.
Takeaway \(\frac{dy}{dx} = f(x)\) plus one point: integrate, then substitute to find \(c\).
4. INT-06
The trapezium rule with \(2\) strips of equal width is used to estimate \(\displaystyle\int_{0}^{2} x^2\,dx\). Which of the following is true?
- AThe estimate is \(\dfrac{8}{3}\) and it is exact.
- BThe estimate is \(3\) and it is an overestimate.
- CThe estimate is \(\dfrac{5}{2}\) and it is an overestimate.
- DThe estimate is \(\dfrac{5}{2}\) and it is an underestimate.
- EThe estimate is \(3\) and it is an underestimate.
Answer: B. \(h = 1\): \(\frac{h}{2}[y_0 + 2y_1 + y_2] = \frac12[0 + 2(1) + 4] = 3\). The true value is \(\frac83\); the curve bends upwards (convex), so the chords lie above it and the rule overestimates.
Full method — how and why
Apply the trapezium rule and then decide over- or under-estimate from the curve's shape.
- Two strips on \([0,2]\) means \(h = 1\) and ordinates at \(x = 0, 1, 2\): \(y = 0, 1, 4\).\(h = \frac{b-a}{n}\) with \(n\) the number of strips (not the number of ordinates).
- Apply the rule: \(\dfrac h2\left[y_0 + 2y_1 + y_2\right] = \dfrac12\left[0 + 2(1) + 4\right] = 3\).Only the interior ordinates are doubled.
- Compare with the exact value \(\int_0^2 x^2 dx = \frac83 \approx 2.67\).The estimate \(3\) is larger.
- Explain it: \(y = x^2\) bends upwards (convex), so each chord lies above the curve and the trapezia overestimate.Convex \(\Rightarrow\) overestimate; concave \(\Rightarrow\) underestimate. This is the reasoning the question wants.
Where it goes wrong
- Forgetting to double the interior ordinate, giving \(\frac52\) - offered twice as a distractor.
- Saying 'underestimate' because the trapezia look like they cut corners - for a convex curve they do the opposite.
Takeaway Trapezium rule: \(\frac h2[y_0 + 2(\text{middles}) + y_n]\); convex curves are overestimated.
5. INT-07
Given that \(\displaystyle\int_{0}^{a} (2x + 1)\,dx = 12\) where \(a > 0\), find \(a\).
- A\(4\)
- B\(12\)
- C\(2\)
- D\(3\)
- E\(\sqrt{12}\)
- F\(6\)
Answer: D. \(a^2 + a = 12 \Rightarrow (a + 4)(a - 3) = 0 \Rightarrow a = 3\).
Full method — how and why
A definite integral with an unknown limit. Integrate, substitute, and solve the resulting equation.
- Integrate: \(\displaystyle\int_0^a (2x+1)dx = \left[x^2 + x\right]_0^a = a^2 + a\).The lower limit \(0\) contributes nothing, which keeps this clean.
- Set equal to \(12\): \(a^2 + a - 12 = 0\).Bring everything to one side.
- Factorise: \((a+4)(a-3) = 0\), so \(a = -4\) or \(a = 3\).Two numbers multiplying to \(-12\) and adding to \(1\).
- Apply the condition \(a > 0\): \(a = 3\).The question's restriction exists to select one root.
Where it goes wrong
- Forgetting the '+x' from integrating the constant \(1\), giving \(a = \sqrt{12}\) - an offered option.
- Giving both roots when only the positive one is allowed.
Takeaway An unknown limit becomes an equation after integration; use any stated sign condition to choose the root.
6. INT-08
Let \(F(x) = \displaystyle\int_{2}^{x} (t^2 - 3t)\,dt\). Find \(F'(3)\).
- A\(9\)
- B\(\dfrac{9}{2}\)
- C\(-3\)
- D\(-\dfrac{7}{6}\)
- E\(0\)
- F\(3\)
Answer: E. By the Fundamental Theorem of Calculus \(F'(x) = x^2 - 3x\), so \(F'(3) = 9 - 9 = 0\).
Full method — how and why
A direct test of the Fundamental Theorem of Calculus: differentiating an integral with a variable upper limit gives back the integrand.
- Apply the FTC: \(\dfrac{d}{dx}\displaystyle\int_2^x g(t)\,dt = g(x)\), so \(F'(x) = x^2 - 3x\).The lower limit is a constant, so it contributes nothing to the derivative.
- Substitute \(x = 3\): \(F'(3) = 9 - 9 = 0\).Just evaluate the integrand at \(3\) - no integration is needed at all.
Where it goes wrong
- Actually integrating and then differentiating - correct but slow, and error-prone.
- Substituting \(t = 3\) into an antiderivative and answering \(-\frac{7}{6}\) or \(\frac92\), both offered.
Takeaway \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\): differentiation undoes integration, so just evaluate the integrand.
7. INT-09
Which of the following statements is/are true?
- I\(\displaystyle\int_{-2}^{2} x^3\,dx = 0\)
- IIThe area between \(y = x^3\) and the \(x\)-axis from \(x = -2\) to \(x = 2\) is \(0\).
- III\(\displaystyle\int_{-2}^{2} x^2\,dx = 2\int_{0}^{2} x^2\,dx\)
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Answer: F. I: odd function over a symmetric interval, so \(0\). II is false: the area is \(2\int_0^2 x^3\,dx = 8\); the integral is \(0\) because the parts cancel. III: even function, true.
Full method — how and why
Three statements distinguishing signed integrals from areas, using odd and even symmetry.
- I: \(x^3\) is odd, so the contributions from \([-2,0]\) and \([0,2]\) cancel exactly: the integral is \(0\). True.For odd \(f\), \(\int_{-a}^{a} f = 0\) always.
- II: the area counts both pieces positively: \(2\int_0^2 x^3 dx = 2 \times 4 = 8 \neq 0\). False.Area and integral differ precisely when the curve crosses the axis - this is the distinction the question is testing.
- III: \(x^2\) is even, so the two halves are equal: \(\int_{-2}^{2} x^2 dx = 2\int_0^2 x^2 dx\). True.For even \(f\), \(\int_{-a}^{a} f = 2\int_0^a f\).
- I and III only.Match to the option list.
Where it goes wrong
- Treating 'integral' and 'area' as synonyms and marking II true.
- Thinking odd symmetry also halves the area.
Takeaway Odd function over a symmetric interval: integral \(0\) but area positive. Even function: integral is twice the half.
8. INT-10
Find \(\displaystyle\int \left(x + \dfrac{1}{x}\right)^2 dx\).
- A\(\dfrac{x^3}{3} + 2x - \dfrac{1}{x} + c\)
- B\(\dfrac{x^3}{3} - \dfrac{1}{x} + c\)
- C\(\dfrac{1}{3}\left(x + \dfrac1x\right)^3 + c\)
- D\(\dfrac{x^3}{3} + 2x + \dfrac{1}{x} + c\)
- E\(\dfrac{x^3}{3} + 2x + c\)
- F\(x^3 + 2x - \dfrac1x + c\)
Answer: A. \(\left(x + \frac1x\right)^2 = x^2 + 2 + x^{-2}\), which integrates to \(\frac{x^3}{3} + 2x - x^{-1} + c\). (\(\frac13\left(x + \frac1x\right)^3 + c\) is a common error: you cannot integrate a power of a bracket that way.)
Full method — how and why
Expand before integrating - there is no rule for integrating a power of a bracket on this specification.
- Expand: \(\left(x + \frac1x\right)^2 = x^2 + 2\left(x\cdot\frac1x\right) + \frac{1}{x^2} = x^2 + 2 + x^{-2}\).The cross term is \(2 \times x \times \frac1x = 2\), a constant - easy to lose.
- Integrate term by term: \(\frac{x^3}{3} + 2x + \frac{x^{-1}}{-1} = \frac{x^3}{3} + 2x - \frac1x\).\(\int x^{-2}dx = \frac{x^{-1}}{-1} = -x^{-1}\); the index \(-2\) is allowed because it is not \(-1\).
- Add the constant: \(\dfrac{x^3}{3} + 2x - \dfrac1x + c\).Indefinite integrals always need \(+c\).
Where it goes wrong
- Choosing \(\frac13\left(x+\frac1x\right)^3 + c\) - you cannot integrate a bracket by raising its power.
- Sign error giving \(+\frac1x\), or dropping the \(2x\) from the cross term.
Takeaway Expand brackets fully first; \(\int x^{-2}dx = -x^{-1}\), and the cross term of a square is easy to forget.
9. INT-12
Find the value of \(\displaystyle\int_{1}^{8} x^{-\frac{2}{3}}\,dx\).
- A\(\dfrac{21}{2}\)
- B\(9\)
- C\(3\)
- D\(6\)
- E\(2\)
- F\(\dfrac{3}{2}\)
Answer: C. \(\left[3x^{1/3}\right]_1^8 = 3(2 - 1) = 3\).
Full method — how and why
A negative fractional index. Add one to it, and choose limits that make the cube root easy.
- Add one to the index: \(-\frac23 + 1 = \frac13\), and divide by \(\frac13\) (i.e. multiply by \(3\)): \(\displaystyle\int x^{-2/3}dx = 3x^{1/3}\).Dividing by a fraction multiplies by its reciprocal.
- Evaluate at the limits: \(3\left(8^{1/3}\right) - 3\left(1^{1/3}\right) = 3(2) - 3(1)\).\(8^{1/3} = 2\) - the limits were chosen to be perfect cubes.
- Subtract: \(6 - 3 = 3\).Simple arithmetic.
Where it goes wrong
- Forgetting the factor \(3\) and answering \(1\) or \(2\).
- Adding one to the index the wrong way and using \(x^{-5/3}\).
Takeaway \(\int x^n dx = \frac{x^{n+1}}{n+1}\) works for negative and fractional \(n\) (except \(n = -1\)); dividing by \(\frac13\) multiplies by \(3\).
10. INT-13
\(f(x) = ax + b\) satisfies \(\displaystyle\int_{0}^{1} f(x)\,dx = 2\) and \(\displaystyle\int_{0}^{2} f(x)\,dx = 6\). Find \(f(1)\).
- A\(4\)
- B\(0\)
- C\(5\)
- D\(1\)
- E\(3\)
- F\(2\)
Answer: E. \(\frac a2 + b = 2\) and \(2a + 2b = 6\), so \(a + b = 3\); subtracting the first equation from this gives \(\frac a2 = 1\), so \(a = 2\), \(b = 1\) and \(f(1) = 3\).
Full method — how and why
Two integral conditions on a linear function give two equations in \(a\) and \(b\).
- Integrate in general: \(\displaystyle\int_0^k (ax+b)dx = \frac{ak^2}{2} + bk\).Doing it once with a general limit saves repeating the work.
- \(k = 1\): \(\frac a2 + b = 2\). \(k = 2\): \(2a + 2b = 6\), i.e. \(a + b = 3\).Substitute each limit and simplify the second by dividing by \(2\).
- Subtract the first from the second: \(\frac a2 = 1\), so \(a = 2\), and then \(b = 1\).Elimination: \(a - \frac a2 = \frac a2\) and \(3 - 2 = 1\).
- Compute \(f(1) = a + b = 3\).Conveniently, \(f(1) = a + b\), which the second equation already gave.
Where it goes wrong
- Forgetting the \(\frac12\) in \(\frac{ak^2}{2}\) and getting a different pair.
- Answering \(2\) (the value of the first integral) instead of \(f(1)\).
Takeaway Integrate once with a general limit, then substitute each condition to get simultaneous equations.
11. INT-16
Given that \(\displaystyle\int_{0}^{4} f(x)\,dx = 10\), find \(\displaystyle\int_{0}^{4} \left(f(x) + 3x\right)dx - \int_{4}^{0} f(x)\,dx\).
- A\(20\)
- B\(0\)
- C\(14\)
- D\(44\)
- E\(24\)
- F\(34\)
Answer: D. \(10 + 24 - (-10) = 44\).
Full method — how and why
Three integral rules in one: linearity, the integral of \(3x\), and the effect of swapping limits.
- Split the first integral: \(\int_0^4 (f(x) + 3x)dx = \int_0^4 f(x)dx + \int_0^4 3x\,dx = 10 + \int_0^4 3x\,dx\).Sums split into separate integrals.
- Compute \(\int_0^4 3x\,dx = \left[\frac{3x^2}{2}\right]_0^4 = 24\).\(\frac{3 \times 16}{2} = 24\).
- Handle the second term: \(\int_4^0 f(x)dx = -\int_0^4 f(x)dx = -10\).Reversing the limits reverses the sign - the key fact here.
- Combine: \((10 + 24) - (-10) = 44\).Subtracting a negative adds.
Where it goes wrong
- Treating \(\int_4^0 f\) as \(+10\) and answering \(24\).
- Forgetting the \(3x\) term, giving \(20\).
Takeaway \(\int_b^a = -\int_a^b\); watch for reversed limits used as a deliberate trap.
12. INT-17
The gradient of a curve at \((x, y)\) is \(3\sqrt{x}\) and the curve passes through \((4, 20)\). Find the \(y\)-intercept of the curve.
- A\(12\)
- B\(20\)
- C\(4\)
- D\(8\)
- E\(-4\)
- F\(0\)
Answer: C. \(y = 2x^{3/2} + c\); \(20 = 16 + c \Rightarrow c = 4\). The \(y\)-intercept is \(4\).
Full method — how and why
Integrate the gradient function, use the given point for \(c\), then read the intercept.
- Integrate: \(\displaystyle\int 3x^{1/2}dx = 3 \times \frac{x^{3/2}}{3/2} = 2x^{3/2}\), so \(y = 2x^{3/2} + c\).\(3 \div \frac32 = 2\).
- Substitute \((4, 20)\): \(4^{3/2} = 8\), so \(2(8) + c = 20\), giving \(c = 4\).\(4^{3/2} = (\sqrt4)^3 = 8\).
- The \(y\)-intercept is the value at \(x = 0\): \(y = 0 + 4 = 4\).For this curve the intercept is just \(c\).
Where it goes wrong
- Using \(4^{3/2} = 6\) and getting \(c = 8\) - an offered option.
- Reporting \(20\) (the given \(y\)-value) as the intercept.
Takeaway The constant of integration is the \(y\)-intercept when the rest of the expression vanishes at \(x = 0\).
13. INT-18
Let \(I = \displaystyle\int_{0}^{1} (x^2 + 1)\,dx\) and \(J = \displaystyle\int_{0}^{1} (x + 1)^2\,dx\). Which of the following is true?
- A\(I < J\) and \(J - I = \dfrac{4}{3}\)
- B\(I < J\) and \(J - I = \dfrac{1}{2}\)
- C\(I = J\)
- D\(I < J\) and \(J - I = 1\)
- E\(I > J\)
Answer: D. \(I = \frac13 + 1 = \frac43\). \(J = \int_0^1 (x^2 + 2x + 1)\,dx = \frac13 + 1 + 1 = \frac73\). \(J - I = 1\).
Full method — how and why
Two similar-looking integrals. Expand the second - \((x+1)^2 \neq x^2 + 1\) - and compare.
- Compute \(I\): \(\displaystyle\int_0^1 (x^2+1)dx = \left[\frac{x^3}{3} + x\right]_0^1 = \frac13 + 1 = \frac43\).Term by term.
- Expand the second integrand: \((x+1)^2 = x^2 + 2x + 1\).The cross term \(2x\) is the whole difference between the two integrals.
- Compute \(J\): \(\left[\frac{x^3}{3} + x^2 + x\right]_0^1 = \frac13 + 1 + 1 = \frac73\).The extra \(\int_0^1 2x\,dx = 1\).
- Compare: \(J - I = \frac73 - \frac43 = 1\), and \(J > I\).Equivalently, the difference is exactly \(\int_0^1 2x\,dx = 1\).
Where it goes wrong
- Assuming \(I = J\) because the integrands 'look the same'.
- Arithmetic giving \(J - I = \frac43\) - an offered option.
Takeaway \((x+1)^2 = x^2 + 2x + 1\): the difference between the two integrals is the integral of the cross term.
14. INT-20
Evaluate \(\displaystyle\int_{-1}^{2} |x|\,dx\).
- A\(2\)
- B\(\dfrac{7}{2}\)
- C\(3\)
- D\(\dfrac{5}{2}\)
- E\(\dfrac{1}{2}\)
- F\(\dfrac{3}{2}\)
Answer: D. Two triangles: \(\frac12\cdot1\cdot1 + \frac12\cdot2\cdot2 = \frac12 + 2 = \frac52\).
Full method — how and why
A modulus integrand. Split at the corner, or - faster - read the areas off as triangles.
- Note \(|x|\) changes formula at \(x = 0\), which is inside \([-1, 2]\).Split at every point where the modulus switches branch.
- On \([-1, 0]\): \(|x| = -x\), and the graph is a triangle with base \(1\) and height \(1\), area \(\frac12\).Sketching \(y = |x|\) makes the integral a pair of triangles - no antiderivative needed.
- On \([0, 2]\): \(|x| = x\), a triangle with base \(2\) and height \(2\), area \(2\).\(\frac12 \times 2 \times 2 = 2\).
- Add: \(\frac12 + 2 = \frac52\).Both pieces are above the axis, so both count positively.
Where it goes wrong
- Integrating \(x\) straight through and getting \(\frac{4}{2} - \frac12 = \frac32\) - an offered option.
- Treating the left piece as negative.
Takeaway \(|x|\) is never negative: split at the corner, and use triangle areas when the graph is piecewise linear.
15. X2-17
The trapezium rule with \(n\) strips is used to estimate \(\displaystyle\int_{0}^{1} \sqrt{x}\,dx\). Which of the following is true for every \(n\)?
- AThe estimate is an overestimate.
- BThe estimate is an underestimate.
- CThe estimate is an underestimate only when \(n\) is even.
- DThe estimate is exact.
- EIt cannot be decided without calculating.
Answer: B. \(y = \sqrt x\) is concave (bends downwards), so every chord lies below the curve and the trapezia underestimate the area.
Full method — how and why
Whether the trapezium rule over- or under-estimates depends on the concavity of the curve.
- The trapezium rule replaces the curve on each strip by the straight chord joining the endpoints.Its estimate is the area under those chords.
- \(y=\sqrt x\) is concave: it rises steeply then flattens, bending downwards.Its second derivative \(-\frac14x^{-3/2}\) is negative for \(x>0\) - or just picture the shape.
- For a concave curve every chord lies below the arc, so each trapezium sits below the true region.That is what concavity means geometrically.
- So the estimate is an underestimate, for every \(n\).The number of strips affects accuracy, not direction.
Where it goes wrong
- Assuming the trapezium rule always overestimates - true only for convex (upward-bending) curves such as \(y=x^2\).
- Thinking the parity of \(n\) matters - that is Simpson's rule, which needs an even number of strips.
- Answering 'cannot be decided' when the shape settles it.
Takeaway Trapezium rule: convex curve \(\Rightarrow\) overestimate, concave curve \(\Rightarrow\) underestimate. Sketch and look at the bend.
16. INT-01
Evaluate \(\displaystyle\int_{1}^{4} \dfrac{3x^2 - 5}{\sqrt{x}}\,dx\).
- A\(\dfrac{146}{5}\)
- B\(\dfrac{186}{5}\)
- C\(28\)
- D\(\dfrac{86}{5}\)
- E\(\dfrac{126}{5}\)
- F\(\dfrac{136}{5}\)
Answer: F. Integrand \(= 3x^{3/2} - 5x^{-1/2}\); integral \(= \left[\frac65 x^{5/2} - 10x^{1/2}\right]_1^4 = \left(\frac{6\cdot32}{5} - 20\right) - \left(\frac65 - 10\right) = \frac{192}{5} - 20 - \frac65 + 10 = \frac{186}{5} - 10 = \frac{136}{5}\).
Full method — how and why
You can only integrate powers of \(x\), so split the fraction first. Then the limits are straightforward - if the fractions are handled carefully.
- Split the fraction: \(\dfrac{3x^2 - 5}{\sqrt x} = 3x^2 x^{-1/2} - 5x^{-1/2} = 3x^{3/2} - 5x^{-1/2}\).Dividing by \(\sqrt x = x^{1/2}\) subtracts \(\frac12\) from each index.
- Integrate: \(\displaystyle\int 3x^{3/2}\,dx = \frac{3x^{5/2}}{5/2} = \frac65 x^{5/2}\) and \(\displaystyle\int 5x^{-1/2}\,dx = 10x^{1/2}\).Add one to the index and divide by the new index: \(\frac32 + 1 = \frac52\), and \(-\frac12 + 1 = \frac12\) so dividing by \(\frac12\) doubles.
- Evaluate at \(x = 4\): \(4^{5/2} = 32\) and \(4^{1/2} = 2\), giving \(\frac{192}{5} - 20\).\(4^{5/2} = (\sqrt4)^5 = 2^5\).
- Evaluate at \(x = 1\): \(\frac65 - 10\). Subtract: \(\left(\frac{192}{5} - 20\right) - \left(\frac65 - 10\right) = \frac{186}{5} - 10 = \frac{136}{5}\).\(\frac{192 - 6}{5} = \frac{186}{5}\), then \(-20 + 10 = -10 = -\frac{50}{5}\).
Where it goes wrong
- Forgetting the lower limit entirely, giving \(\frac{186}{5}\) - an offered option.
- Dividing by \(\frac52\) incorrectly: \(\frac{3}{5/2} = \frac65\), not \(\frac{15}{2}\).
Takeaway Split fractions into powers of \(x\) before integrating; \(4^{5/2} = 32\) type evaluations are worth doing slowly.
17. INT-05
Find the area of the finite region enclosed by the curve \(y = x^2\) and the line \(y = 2x + 3\).
- A\(\dfrac{16}{3}\)
- B\(\dfrac{28}{3}\)
- C\(12\)
- D\(\dfrac{20}{3}\)
- E\(\dfrac{32}{3}\)
- F\(\dfrac{64}{3}\)
Answer: E. Intersections: \(x^2 - 2x - 3 = (x - 3)(x + 1) = 0\). Area \(= \int_{-1}^{3} (2x + 3 - x^2)\,dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3} = (9 + 9 - 9) - \left(1 - 3 + \frac13\right) = 9 + \frac53 = \frac{32}{3}\).
Full method — how and why
Area between a line and a curve: integrate (upper \(-\) lower) between the intersection points.
- Find the intersections: \(x^2 = 2x + 3 \Rightarrow x^2 - 2x - 3 = (x-3)(x+1) = 0\), so \(x = -1\) and \(x = 3\).These are the limits of the enclosed region.
- Decide which is on top: at \(x = 0\) the line gives \(3\) and the curve gives \(0\), so the line is above.Testing one interior point settles the order - guessing wrong gives the negative of the answer.
- Integrate the difference: \(\displaystyle\int_{-1}^{3}\left(2x + 3 - x^2\right)dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3}\).Subtracting first, then integrating once, is quicker and less error-prone than two separate integrals.
- Evaluate: at \(3\), \(9 + 9 - 9 = 9\); at \(-1\), \(1 - 3 + \frac13 = -\frac53\). Difference: \(9 + \frac53 = \frac{32}{3}\).\(-\frac{(-1)^3}{3} = +\frac13\) - the double negative is the classic slip.
Where it goes wrong
- Integrating (curve \(-\) line) and giving \(-\frac{32}{3}\), or taking the magnitude of the wrong piece.
- Sign error at the lower limit, giving \(\frac{16}{3}\) or \(\frac{28}{3}\).
Takeaway Area between curves \(= \int(\text{upper} - \text{lower})\) between intersections; check which is upper with one test value.
18. INT-11
The region bounded by \(y = 4 - x^2\), the \(x\)-axis and the lines \(x = 0\) and \(x = 3\) is shaded. Find the total shaded area.
- A\(3\)
- B\(\dfrac{7}{3}\)
- C\(\dfrac{16}{3}\)
- D\(\dfrac{25}{3}\)
- E\(\dfrac{23}{3}\)
- F\(9\)
Answer: E. From \(0\) to \(2\) the curve is above the axis: \(\int_0^2 (4 - x^2)\,dx = 8 - \frac83 = \frac{16}{3}\). From \(2\) to \(3\) it is below: \(\left|\int_2^3 (4 - x^2)\,dx\right| = \left|(12 - 9) - \frac{16}{3}\right| = \frac73\). Total \(\frac{23}{3}\).
Full method — how and why
The curve crosses the axis inside the range, so the region must be split - otherwise the pieces cancel.
- Find where the curve meets the axis: \(4 - x^2 = 0\) at \(x = \pm2\), so \(x = 2\) is inside \([0,3]\).A crossing inside the limits is the signal to split the integral.
- From \(0\) to \(2\) the curve is above the axis: \(\int_0^2 (4 - x^2)dx = \left[4x - \frac{x^3}{3}\right]_0^2 = 8 - \frac83 = \frac{16}{3}\).Positive, as expected.
- From \(2\) to \(3\) it is below: \(\int_2^3 (4-x^2)dx = \left(12 - 9\right) - \left(8 - \frac83\right) = 3 - \frac{16}{3} = -\frac73\).Negative, so take its magnitude \(\frac73\) for the area.
- Add the magnitudes: \(\dfrac{16}{3} + \dfrac{7}{3} = \dfrac{23}{3}\).Total shaded area, both pieces counted positively.
Where it goes wrong
- Integrating straight through from \(0\) to \(3\) and getting \(3\) - which is offered.
- Forgetting to take the magnitude of the second piece.
Takeaway Split at every crossing inside the limits and add the magnitudes; a single integral would cancel the pieces.
19. INT-14
The trapezium rule is used to estimate \(\displaystyle\int_{a}^{b} f(x)\,dx\). For which of the following functions is the estimate guaranteed to be an underestimate for every choice of \(a < b\)?
- A\(f(x) = -x^2\)
- B\(f(x) = x^4\)
- C\(f(x) = 2x + 1\)
- D\(f(x) = x^2\)
- E\(f(x) = x^3\)
Answer: A. The trapezium rule underestimates when the curve is concave (bends downwards) so that chords lie below the curve: \(y = -x^2\) everywhere. \(x^3\) changes concavity at \(0\).
Full method — how and why
A conceptual question about the trapezium rule: it is the curve's concavity, not its sign or steepness, that decides.
- Recall the rule: the trapezium rule replaces the curve by straight chords.Each strip's true area is compared with a trapezium whose top is the chord.
- A chord lies below a concave (downward-bending) curve, so the estimate is too small: an underestimate.Concave means \(f'' < 0\) throughout - the curve bulges above its chords.
- Check the options: \(-x^2\) has \(f'' = -2 < 0\) everywhere ✓. \(x^2\) and \(x^4\) are convex (overestimate); \(2x+1\) is straight (exact); \(x^3\) changes concavity at \(0\), so it depends on the interval.'Guaranteed for every choice of \(a < b\)' rules out \(x^3\), whose behaviour varies.
- So \(f(x) = -x^2\).The only universally concave option.
Where it goes wrong
- Choosing \(x^3\) without noticing that it is concave for \(x<0\) but convex for \(x>0\).
- Thinking a negative function must give an underestimate - sign is irrelevant; shape is everything.
Takeaway Trapezium rule: concave (\(f''<0\)) underestimates, convex (\(f''>0\)) overestimates, linear is exact.
20. INT-15
The area between the curve \(y = x^2\) and the line \(y = k\) (where \(k > 0\)) is \(\dfrac{32}{3}\). Find \(k\).
- A\(8\)
- B\(3\)
- C\(\sqrt{8}\)
- D\(4\)
- E\(2\)
- F\(16\)
Answer: D. Area \(= \int_{-\sqrt k}^{\sqrt k}(k - x^2)\,dx = 2\left(k^{3/2} - \frac{k^{3/2}}{3}\right) = \frac43 k^{3/2} = \frac{32}{3} \Rightarrow k^{3/2} = 8 \Rightarrow k = 4\).
Full method — how and why
Work backwards from a given area to the parameter. Symmetry halves the work.
- Find the intersections: \(x^2 = k\) at \(x = \pm\sqrt k\).The horizontal line cuts the parabola at two symmetric points.
- Set up the area: \(\displaystyle\int_{-\sqrt k}^{\sqrt k}\left(k - x^2\right)dx\), with the line above the curve between the roots.Upper minus lower.
- Use symmetry: \(= 2\displaystyle\int_0^{\sqrt k}(k - x^2)dx = 2\left[kx - \frac{x^3}{3}\right]_0^{\sqrt k} = 2\left(k^{3/2} - \frac{k^{3/2}}{3}\right) = \frac43 k^{3/2}\).The integrand is even, so integrate over half the range and double. Note \(k\sqrt k = k^{3/2}\).
- Solve \(\frac43 k^{3/2} = \frac{32}{3}\): \(k^{3/2} = 8\), so \(k = 8^{2/3} = 4\).Undo the \(\frac32\) power by raising to the \(\frac23\): \(8^{2/3} = (\sqrt[3]8)^2 = 4\).
Where it goes wrong
- Solving \(k^{3/2} = 8\) as \(k = 8^{3/2} = 22.6\) or \(k = \sqrt8\) - both offered in spirit.
- Forgetting to double after using symmetry.
Takeaway Use even symmetry to halve the integration; undo a \(\frac32\) power by raising to \(\frac23\).
21. INT-19
The curve \(y = x(x - 1)(x - 2)\) and the \(x\)-axis enclose two regions. What is the ratio of the area of the region for \(0 \leq x \leq 1\) to the area for \(1 \leq x \leq 2\)?
- A\(1 : 2\)
- B\(2 : 1\)
- C\(1 : 3\)
- D\(1 : 4\)
- E\(1 : 1\)
- F\(3 : 1\)
Answer: E. The cubic has rotational symmetry about \((1, 0)\) (it is \(u^3 - u\) with \(u = x - 1\), an odd function), so the two regions are congruent: ratio \(1 : 1\).
Full method — how and why
A symmetry question. Shifting the cubic reveals it is odd about \((1,0)\), so the two regions are congruent - no integration needed.
- Substitute \(u = x - 1\) to centre the cubic: \(y = (u+1)u(u-1) = u(u^2 - 1) = u^3 - u\).The roots \(0, 1, 2\) become \(-1, 0, 1\) - symmetric about the origin, which is the hint to try this.
- \(u^3 - u\) is an odd function: replacing \(u\) by \(-u\) negates it.Odd functions have rotational symmetry of order 2 about the origin.
- So the piece on \(-1 \leq u \leq 0\) is the \(180^\circ\) rotation of the piece on \(0 \leq u \leq 1\): the regions are congruent and have equal areas.Rotation preserves area.
- The ratio is \(1 : 1\).You could integrate both (each has area \(\frac14\)), but the symmetry argument is instant.
Where it goes wrong
- Assuming the regions differ because one is above the axis and one below - sign differs, area does not.
- Integrating carelessly and comparing signed values.
Takeaway Shift a cubic so its middle root is at the origin: if it becomes odd, the two enclosed regions are congruent.
22. X2-03
\(f\) is a function defined for all real \(x\) and \(\displaystyle\int_{0}^{4} f(x)\,dx = 0\). Which of the following statements must be true?
- I\(f(x) = 0\) for some \(x\) with \(0 \leq x \leq 4\).
- IIThe area between the curve and the \(x\)-axis from \(0\) to \(4\) is \(0\).
- III\(\displaystyle\int_{0}^{2} f(x)\,dx = -\int_{2}^{4} f(x)\,dx\)
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Answer: D. III is just the rule for combining integrals over adjacent ranges. I fails for a function that jumps from \(-1\) to \(1\) without passing through \(0\) (nothing says \(f\) is continuous). II fails: \(f(x) = x - 2\) has integral \(0\) but area \(4\).
Full method — how and why
A hard 'must be true' about integrals. Two claims import assumptions the question never makes.
- III: \(\int_0^4 = \int_0^2 + \int_2^4\), and the left side is \(0\), so \(\int_0^2 = -\int_2^4\). True - pure algebra of integrals.Splitting an integral at an interior point is always valid where the integral exists.
- I: 'zero integral \(\Rightarrow\) \(f\) hits zero' is the intermediate value theorem in disguise - but that needs continuity, which is not given.Read the hypotheses: it only says \(f\) is defined for all real \(x\).
- Counterexample for I: a step function equal to \(-1\) on \([0,2)\) and \(1\) on \([2,4]\) has integral \(0\) but never equals \(0\). False.It jumps straight past zero.
- II: 'integral \(0\)' means signed areas cancel, not that the area is \(0\). Take \(f(x) = x-2\): \(\int_0^4 (x-2)dx = 0\), but the area is \(2 + 2 = 4\). False.Area counts the region below the axis as positive; the integral counts it as negative.
- So III only.Match to the list.
Where it goes wrong
- Accepting I by silently assuming \(f\) is continuous - an assumption the question deliberately withholds.
- Treating 'integral' and 'area' as the same thing; they agree only when \(f \geq 0\) throughout.
Takeaway Integral \(\neq\) area when the function changes sign; and the intermediate value theorem requires continuity, which must be given.