Integration, definite integrals and areas – IB Maths exam questions with mark scheme

1CoreNon-calculator4 marks

A curve $y=f(x)$ has gradient function $f'(x)=6x^2-4x+1$ and passes through the point $(1,5)$. Find $f(x)$.

Attempts to integrate (at least one power increased correctly)M1
$2x^3-2x^2+x$ (with or without $+c$)A1
Substitutes $x=1$, $f=5$: $2-2+1+c=5$M1
$f(x)=2x^3-2x^2+x+4$A1
Answer: $f(x)=2x^3-2x^2+x+4$
Examiner tip: Include $+c$ at the integration stage; without it the substitution mark cannot be earned.

2StandardNon-calculator4 marks

Find the exact value of $\displaystyle\int_0^2\dfrac{4x}{x^2+1}\,dx$.

Recognises the form $\dfrac{g'(x)}{g(x)}$ with $g=x^2+1$ (or substitution $u=x^2+1$)M1
$2\ln(x^2+1)$A1
Substitutes limits: $2\ln5-2\ln1$M1
$2\ln 5$ (accept $\ln25$)A1
Answer: $2\ln5$
Examiner tip: The numerator is exactly twice the derivative of the denominator, so the factor 2 must be kept outside the logarithm.

3StandardNon-calculator5 marks

The diagram shows the curves $y=6x-x^2$ and $y=x^2$. The curves intersect at the origin and at a second point. The shaded region $R$ is bounded by the two curves.

Find the area of $R$.

Equates the curves: $6x-x^2=x^2$M1
$x=0$ and $x=3$A1
Area $=\int_0^3\big[(6x-x^2)-x^2\big]\,dx$ (upper minus lower, correct limits)M1
Antiderivative $3x^2-\dfrac{2x^3}{3}$A1
Area $=27-18=9$A1
Answer: Area $=9$ square units
Examiner tip: Always subtract lower from upper; integrating each curve separately and mixing up the order gives a negative or wrong area.

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4HardNon-calculator6 marks

Let $R_k$ be the region bounded by the curve $y=2xe^{-x^2}$, the $x$-axis and the line $x=k$, where $k>0$.

(a) Find the value of $k$ for which the area of $R_k$ is $\dfrac12$. Give your answer in the form $\sqrt{\ln a}$. [5]

(b) Write down the limit of the area of $R_k$ as $k\to\infty$. [1]

Area $=\int_0^k2xe^{-x^2}dx$ (by inspection or $u=-x^2$)M1
$\big[-e^{-x^2}\big]_0^k$A1
$=1-e^{-k^2}$A1
Sets $1-e^{-k^2}=\dfrac12$, so $e^{-k^2}=\dfrac12$, $k^2=\ln2$M1
$k=\sqrt{\ln 2}\ (\approx0.833)$, rejecting the negative rootA1
Limit $=1$ (since $e^{-k^2}\to0$)B1
Answer: (a) $k=\sqrt{\ln2}$; (b) $1$
Examiner tip: Differentiate your antiderivative as a check: $\frac{d}{dx}(-e^{-x^2})=2xe^{-x^2}$ confirms the sign.

5CoreNon-calculator5 marks

(a) Find $\displaystyle\int\big(4\cos 2x-3e^{-x}\big)\,dx$. [3]

(b) Find $\displaystyle\int\dfrac{6}{3x-1}\,dx$, for $x>\dfrac13$. [2]

(a) $2\sin 2x$ (accept $\dfrac{4}{2}\sin 2x$)A1
(a) $+3e^{-x}$ (sign must be correct)A1
(a) $+\,c$ included in (a). Penalise a missing $+\,c$ once only: if it is missing in (a), condone its omission in (b)A1
(b) Recognises a logarithm of the linear function: $k\ln(3x-1)$ for any non-zero $k$ (accept $k\ln|3x-1|$)M1
(b) $2\ln(3x-1)+c$ (accept $2\ln|3x-1|+c$ or $\ln\big(C(3x-1)^2\big)$)A1
Answer: (a) $2\sin 2x+3e^{-x}+c$ (b) $2\ln(3x-1)+c$
Examiner tip: With a linear inside function, divide by the coefficient of $x$: $\int\cos 2x\,dx=\tfrac12\sin 2x$, not $2\sin 2x$; and $\int e^{-x}dx=-e^{-x}$, so the sign flips.

6StandardNon-calculator6 marks

A function $f$ is continuous for all $x$, and $\displaystyle\int_1^4 f(x)\,dx=7$.

(a) Find $\displaystyle\int_1^4\big(2f(x)-3\big)\,dx$. [3]

(b) Write down the value of $\displaystyle\int_4^1 f(x)\,dx$. [1]

(a) Splits the integral: $2\int_1^4 f(x)\,dx-\int_1^4 3\,dx$ (accept $2\times7-\ldots$)M1
(a) $\int_1^4 3\,dx=9$ (accept $[3x]_1^4$ evaluated as $12-3$)A1
(a) $14-9=5$A1
(b) $-7$A1
(c) Recognises a horizontal translation by 1 unit left (or substitution $u=x+1$ giving limits $1$ and $4$): $\int_0^3 f(x+1)\,dx=\int_1^4 f(u)\,du$R1
(c) $7$ (do not award A1 for 7 with no reasoning)A1
Answer: (a) $5$ (b) $-7$ (c) $7$
Examiner tip: In (a) the constant must be integrated too: $\int_1^4 3\,dx=9$, not $3$. A common wrong answer is $14-3=11$.

7StandardCalculator7 marks

Let $f(x)=2\ln(x+1)-x$, for $0\le x\le5$. The graph of $f$ meets the $x$-axis at the origin and at $x=p$. Region $A$ lies above the $x$-axis between $x=0$ and $x=p$; region $B$ is bounded by the graph of $f$, the $x$-axis and the line $x=5$, as shown in the diagram.

(a) Find the value of $p$. [2]

(b) Find the total area of the shaded regions $A$ and $B$. [3]

(a) Attempts to solve $2\ln(x+1)-x=0$ using technology, e.g. graph intercept or equation solverM1
(a) $p=2.51$ (2.51286…)A1
(b) Correct method for total area: $\int_0^p f\,dx-\int_p^5 f\,dx$ or $\int_0^5|f(x)|\,dx$M1
(b) Area of $A=\int_0^p f(x)\,dx=0.644$ (0.64437…) or area of $B=\left|\int_p^5 f(x)\,dx\right|=1.64$ (1.64326…)A1
(b) Total area $=2.29$ (2.28763…)A1
(c) $\int_0^5 f(x)\,dx=-0.999$ (−0.99888…)A1
(c) Explanation: region $B$ is below the $x$-axis so its integral is negative and cancels part of region $A$ (accept: the integral gives signed area, $0.644-1.64$)R1
Answer: (a) $p=2.51$ (b) $2.29$ (c) $-0.999$; the integral over $B$ is negative and is subtracted rather than added
Examiner tip: Typing $\int_0^5 f(x)\,dx$ straight into the GDC gives a signed value; for an area you must split at $x=p$ or integrate $|f(x)|$.

8HardNon-calculator8 marks

The diagram shows the curves $y=\sin 2x$ and $y=\sin x$ for $0\le x\le\pi$. The curves enclose the two shaded regions.

(a) Show that the curves intersect when $x=0$, $x=\dfrac{\pi}{3}$ and $x=\pi$. [3]

(b) Find the total area of the shaded regions. [5]

(a) Uses $\sin 2x=2\sin x\cos x$ and rearranges: $\sin x(2\cos x-1)=0$ (do not award if the equation is divided by $\sin x$ without considering $\sin x=0$)M1
(a) $\sin x=0\Rightarrow x=0,\ \pi$A1
(a) $\cos x=\dfrac12\Rightarrow x=\dfrac{\pi}{3}$ (AG)A1
(b) Sets up two integrals with upper minus lower in each interval: $\int_0^{\pi/3}(\sin 2x-\sin x)\,dx+\int_{\pi/3}^{\pi}(\sin x-\sin 2x)\,dx$ (accept modulus of each integral)M1
(b) Correct antiderivative $-\tfrac12\cos 2x+\cos x$ (or its negative)A1
(b) Substitutes limits into at least one integralM1
(b) $\dfrac14$ and $\dfrac94$A1
(b) Total area $=\dfrac52$ (accept 2.5)A1
Answer: (b) $\dfrac14+\dfrac94=\dfrac52$
Examiner tip: Integrating $\sin 2x-\sin x$ once from $0$ to $\pi$ gives $-2$, not the area: the curves cross at $\pi/3$, so you must split there.

9StandardNon-calculator5 marks

Use the substitution $u=x^2+1$ to find the exact value of $\displaystyle\int_0^{\sqrt3} x\sqrt{x^2+1}\,dx$.

$\dfrac{du}{dx}=2x$, so $x\,dx=\dfrac12\,du$M1
Integral in terms of $u$: $\dfrac12\int u^{1/2}\,du$A1
Correct new limits $u=1$ and $u=4$ (or substitutes back to $x$ and uses original limits)A1
Integrates: $\dfrac12\times\dfrac23u^{3/2}=\dfrac13u^{3/2}$ (accept by inspection $\dfrac13(x^2+1)^{3/2}$)M1
$\dfrac13(8-1)=\dfrac73$A1
Answer: $\dfrac73$
Examiner tip: Change the limits when you change the variable: using $0$ and $\sqrt3$ with $u^{3/2}$ gives the wrong answer.

10StretchNon-calculator7 marks

The diagram shows the curve $y=4x-x^2$ and the line $y=kx$, where $0

(a) Show that the area of $R$ is $\dfrac{(4-k)^3}{6}$. [5]

(b) The line divides the region enclosed by the curve and the $x$-axis into two parts of equal area. Find the exact value of $k$. [2]

(a) Solves $4x-x^2=kx$ to obtain $x=0$ and $x=4-k$A1
(a) Area $=\int_0^{4-k}\big(4x-x^2-kx\big)\,dx$ (curve minus line, correct limits)M1
(a) $\left[\dfrac{(4-k)x^2}{2}-\dfrac{x^3}{3}\right]_0^{4-k}$A1
(a) Substitutes the upper limit: $\dfrac{(4-k)^3}{2}-\dfrac{(4-k)^3}{3}$M1
(a) $=\dfrac{(4-k)^3}{6}$ (AG); must not expand and re-factorise without workingA1
(b) Area under curve $\int_0^4(4x-x^2)\,dx=\dfrac{32}{3}$ and sets $\dfrac{(4-k)^3}{6}=\dfrac{16}{3}$M1
(b) $(4-k)^3=32\Rightarrow k=4-\sqrt[3]{32}$ (accept $4-2\sqrt[3]{4}$; do not accept $0.825$ alone)A1
Answer: (b) $k=4-\sqrt[3]{32}=4-2\sqrt[3]4\ (\approx0.825)$
Examiner tip: Keep $(4-k)$ as a single factor when substituting the upper limit; expanding $(4-k)^3$ early makes the algebra very hard to finish.

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