Integration (definite, areas) – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

Find $$\int\left(6x^{2}-\frac{4}{x^{3}}+3\right)\mathrm{d}x,$$ giving each term in its simplest form. [4]

for $x^{n}\to x^{n+1}$ for at least one termM1
two of $2x^{3},\ +2x^{-2},\ +3x$ correctA1
all three terms correctA1
fully correct with $+c$A1
Answer: $2x^{3}+\dfrac{2}{x^{2}}+3x+c$
Examiner tip: Don't forget '$+c$' on an indefinite integral – it is often worth a whole mark.

2StandardCalculator5 marks

A curve with equation $y=f(x)$, $x>0$, passes through the point $(4,50)$. Given that $$f'(x)=3x^{2}-\frac{2}{\sqrt{x}},$$ find $f(x)$. [5]

for integrating, $x^{n}\to x^{n+1}$ for at least one termM1
$x^{3}$A1
$-4x^{\frac12}$A1
for substituting $x=4,\ y=50$ into their $f(x)$ with $+c$ and solving for $c$M1
$f(x)=x^{3}-4\sqrt{x}-6$A1
Answer: $f(x)=x^{3}-4\sqrt{x}-6$
Examiner tip: $\int -2x^{-\frac12}\,\mathrm{d}x=\dfrac{-2x^{\frac12}}{\frac12}=-4x^{\frac12}$ – dividing by a half doubles.

3StandardCalculator4 marks

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Find the exact value of $$\int_{1}^{4}\left(3\sqrt{x}-\frac{2}{x^{2}}\right)\mathrm{d}x.$$ [4]

for integrating, $x^{n}\to x^{n+1}$ for at least one termM1
$2x^{\frac32}+2x^{-1}$A1
for substituting both limits into their integrated function and subtracting the right way roundM1
$\tfrac{25}{2}$A1
Answer: $\tfrac{25}{2}$
Examiner tip: Show the substitution: $\left(16+\tfrac12\right)-(2+2)$ – a bare answer from a calculator scores nothing here.

4StandardCalculator5 marks

The diagram shows part of the curve with equation $y=(x+1)(3-x)$. The curve meets the $x$-axis at the points $A$ and $B$. The region $R$, shown shaded, is bounded by the curve and the $x$-axis.

(a) Write down the $x$-coordinates of $A$ and $B$. [1]

(b) Use calculus to find the exact area of $R$. [4]

(a) $-1$ and $3$ (both required)B1
(b) for expanding to $3+2x-x^{2}$ and integrating, $x^{n}\to x^{n+1}$M1
(b) $3x+x^{2}-\tfrac13x^{3}$A1
(b) for substituting their limits $3$ and $-1$ and subtractingM1
(b) $\tfrac{32}{3}$A1
Answer: (a) $-1,\ 3$ (b) $\tfrac{32}{3}$
Examiner tip: Expand the brackets before integrating – you can't integrate a product factor by factor.

5HardCalculator6 marks

The curve $C$ has equation $y=x^{2}-2x+3$ and the line $l$ has equation $y=x+3$. The region $R$, shown shaded in the diagram, is bounded by $C$ and $l$.

Use algebraic integration to find the exact area of $R$. [6]

for setting $x^{2}-2x+3=x+3$ and solvingM1
$x=0$ and $x=3$A1
for $\int(\text{line}-\text{curve})\,\mathrm{d}x=\int(3x-x^{2})\,\mathrm{d}x$, or area of trapezium minus area under curve (allow curve $-$ line if the sign is dealt with at the end)M1
$\tfrac32x^{2}-\tfrac13x^{3}$ (or $\tfrac13x^{3}-x^{2}+3x$ for the curve)A1
for substituting their limits correctly (and, if using the trapezium method, subtracting from trapezium area $13.5$)dM1
$\tfrac92$A1
Answer: $\tfrac{9}{2}$
Examiner tip: Subtract curve from line (top minus bottom) before integrating – it gives one integral instead of two.

6StandardCalculator6 marks

The diagram shows a sketch of the curve with equation $y=x(x-2)(x-4)$. The regions $R_1$ and $R_2$, shown shaded, are bounded by the curve and the $x$-axis.

(a) Use algebraic integration to find the total area of the shaded regions. [5]

(b) Explain why $\displaystyle\int_{0}^{4}x(x-2)(x-4)\,\mathrm{d}x$ does not give this total area. [1]

(a) for expanding to $x^{3}-6x^{2}+8x$M1
(a) for integrating, $x^{n}\to x^{n+1}$M1
(a) $\tfrac14x^{4}-2x^{3}+4x^{2}$A1
(a) for integrating separately over $[0,2]$ and $[2,4]$ (values $4$ and $-4$) and adding magnitudes (or uses symmetry: $2\times\int_0^2$)M1
(a) total area $=8$A1
(b) $R_2$ lies below the $x$-axis so its integral is negative ($-4$) and cancels with $R_1$ (the integral gives $0$). Accept any equivalent, e.g. 'areas above and below the axis cancel'; do not accept just 'the answer is 0' without reasonB1
Answer: (a) $8$ (b) the integral is $4+(-4)=0$ as $R_2$ lies below the axis
Examiner tip: Split the integral at every root – a region below the $x$-axis gives a negative integral.

7StandardCalculator4 marks

Given that $k>1$ and $$\int_{1}^{k}(2x-3)\,\mathrm{d}x=6,$$ find the value of $k$. [4]

for integrating to $x^{2}-3x$M1
for substituting the limits: $(k^{2}-3k)-(1-3)=6$M1
$k^{2}-3k-4=0$A1
$k=4$ only (with $k=-1$ rejected since $k>1$)A1
Answer: $k=4$
Examiner tip: Subtract the lower-limit value carefully: $-(1-3)=+2$, which is where most sign errors creep in.

8StretchCalculator6 marks

The curve $C$ has equation $y=x^{2}-4x+7$. The point $P(3,4)$ lies on $C$. The line $l$ is the tangent to $C$ at $P$.

The region $R$, shown shaded in the diagram, is bounded by $C$, the line $l$ and the $y$-axis.

Show that the area of $R$ is $9$. [6]

for differentiating: $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x-4$ and substituting $x=3$M1
tangent $l$: $y=2x-2$A1
for $\int(\text{curve}-\text{line})\,\mathrm{d}x=\int(x^{2}-6x+9)\,\mathrm{d}x$ (or a valid area-under-curve minus/plus triangle/trapezium method)M1
$\tfrac13x^{3}-3x^{2}+9x$A1
for using limits $0$ and $3$dM1
area $=9-0=9$ with no errorsA1*
Answer: Area $=\int_{0}^{3}(x-3)^{2}\,\mathrm{d}x=9$
Examiner tip: Find the tangent first; then the region runs from the $y$-axis ($x=0$) to $P$ ($x=3$) – don't stop at where $l$ crosses the $x$-axis.

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