Integration, areas, volumes of revolution (HL), kinematics – 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.

4HardNon-calculator8 marks

A particle moves in a straight line. Its velocity at time $t$ seconds, $t\ge0$, is $v=3t^2-12t+9\ \text{m s}^{-1}$. When $t=0$ the displacement of the particle from the origin is $2$ m.

(a) Find the times at which the particle is instantaneously at rest. [2]

(b) Find an expression for the displacement $s$ of the particle at time $t$. [3]

Sets $3t^2-12t+9=0$ (or $t^2-4t+3=0$)M1
$t=1$ and $t=3$A1
Integrates $v$ (attempt)M1
$s=t^3-6t^2+9t+c$A1
Uses $s(0)=2$: $s=t^3-6t^2+9t+2$A1
Identifies direction changes at $t=1,3$ and evaluates $s$ at $t=0,1,3,4$ (or splits $\int_0^4|v|\,dt$ at $t=1$ and $t=3$)M1
$s=2,\ 6,\ 2,\ 6$ respectively (or the three part-integrals $4,\,-4,\,4$)A1
Distance $=4+4+4=12$ mA1
Answer: (a) $t=1,\,3$; (b) $s=t^3-6t^2+9t+2$; (c) $12$ m
Examiner tip: Displacement from $s(4)-s(0)$ is 4 m, not 12 m; split the motion wherever $v$ changes sign.

5HardNon-calculatorHL only4 marks

The region bounded by the curve $y=\sin x$ for $0\le x\le\pi$ and the $x$-axis is rotated through $360^\circ$ about the $x$-axis. Find the exact volume of the solid formed.

$V=\pi\int_0^{\pi}\sin^2x\,dx$ (limits and $\pi$ present)M1
Uses $\sin^2x=\dfrac{1-\cos2x}{2}$M1
$\dfrac{\pi}{2}\Big[x-\dfrac{\sin2x}{2}\Big]_0^{\pi}$A1
$V=\dfrac{\pi^2}{2}$A1
Answer: $V=\dfrac{\pi^2}{2}$
Examiner tip: Square $y$ before integrating: $\int\sin^2x\,dx$ cannot be done without the double-angle identity.

6HardNon-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.

7HardNon-calculatorHL only4 marks

The region enclosed by the curve $y=x^2$ and the line $y=2x$ is rotated through $360^\circ$ about the $x$-axis. Find the exact volume of the solid generated.

Solves $x^2=2x$ to find limits $x=0$ and $x=2$M1
$V=\pi\int_0^2\big[(2x)^2-(x^2)^2\big]dx$ (difference of squares, not square of difference)M1
Antiderivative $\dfrac{4x^3}{3}-\dfrac{x^5}{5}$A1
$V=\dfrac{64\pi}{15}$A1
Answer: $V=\dfrac{64\pi}{15}$
Examiner tip: Subtract the squares of the radii ($\pi\int(R^2-r^2)$); squaring $(2x-x^2)$ is the classic error.

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