What is 8.3: Definite integrals and areas?

_a^b f(x)\,dx=F(b)-F(a) where F is an antiderivative; write the antiderivative in square brackets with the limits, substitute both, and subtract in the right order, using e^0=1 and 1=0 where they appear.

How many marks are 8.3 questions worth?

8.3 practice drills on this page are worth 3–6 marks each; marks in a real exam vary by question.

Which past exam papers has 8.3 appeared in?

We have not yet indexed a verified past-paper appearance for 8.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

8.3

Evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves.

Draft — not yet indexed

Definite integrals and areas

Worked answers and methods for 8.3 on Edexcel A-level Maths 9MA0.

Definition

Specification 8.3: evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves.

Explanation

$\int_a^b f(x)\,dx=F(b)-F(a)$ where $F$ is an antiderivative; write the antiderivative in square brackets with the limits, substitute both, and subtract in the right order, using $e^0=1$ and $\ln1=0$ where they appear.
The integral is a signed area, so a region below the $x$ -axis integrates to a negative number and its area is the modulus; split at the roots when a region crosses the axis.
For the area between a curve and a line or a second curve, find the intersections first, then integrate $y_{\text{upper}}-y_{\text{lower}}$ between them, or subtract the area of a triangle or trapezium under a straight line from the area under the curve.
Exact answers keep surds, $\pi$ , $e$ and logarithms; “show all stages of working” means calculator-only answers score nothing.
A parametric curve has area $\int y\,\dfrac{dx}{dt}\,dt$ with the limits in $t$ .

For a parametric curve, vertical strips give area

\int y(t)\,\dfrac{dx}{dt}\,dt

Worked example

The curve

C

has equation

y=4x-x^2

and the line

l

has equation

y=x

. The finite region

R

is bounded by

C

and

l

. Use algebraic integration to find the exact area of

R

(5)

(Total for Question 1 is 5 marks)

Working

1.Intersections: $4x-x^2=x$ , so $x(3-x)=0$ and $x=0$ or $x=3$ ; between them the curve is above the line.
2.Area $=\displaystyle\int_0^3\left[(4x-x^2)-x\right]dx=\int_0^3(3x-x^2)\,dx$ .
3. $=\left[\dfrac{3x^2}{2}-\dfrac{x^3}{3}\right]_0^3=\left(\dfrac{27}{2}-9\right)-0=\dfrac92$ .

Answer: $\dfrac{9}{2}$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Sets $4x-x^2=x$ and solves to find the intersections	M1
	$x=0$ and $x=3$	A1
	Integrates the difference of the two functions, $x^n\to x^{n+1}$	M1
	$\dfrac{3x^2}{2}-\dfrac{x^3}{3}$ or $2x^2-\dfrac{x^3}{3}$ and $\dfrac92$	A1
	Applies the limits $0$ and $3$ correctly to obtain $\dfrac92$	A1
Notes
M1: or area under the curve minus the triangle $\frac12\times3\times3$
A1: cso

(5 marks)

Common mistakes

Don't use one pair of limits after the upper and lower curves swap, causing part of an enclosed area to cancel.
Don't integrate one curve from the axis instead of subtracting lower curve from upper curve over the intersection limits.

Mark-scheme wording to use

“y = 8x, 12x + y = 48 ⇒ y = 19.2 (or x = 2.4) M1 Triangle area is (1/2) × 4 × "19.2" (= 38.4 or 192/5) or ∫_0^"2.4" 8x dx + ∫_"2.4"^4 "(48 − 12x)" dx dM1 … M1: Attempts to find either the x or y coordinate of the intersection of line l_1 and line l_2 … dM1: Correct method for the area of the triangle.” — Pearson 9MA0, June 2024 Paper 1, Q10(c)
“M1: Substitutes the limits 1 and 0 into an expression of the form ±αx²e^(−3x) ± βxe^(−3x) ± γe^(−3x), α, β, γ ≠ 0 and subtracts the right way round. … Must see evidence of the use of both limits and subtraction and use of e⁰ = 1. … A1: Correct answer of 16/27 − (136/27)e⁻³ but allow equivalent exact fractions and condone 16/27 − 136/(27e³).” — Pearson 9MA0, June 2024 Paper 2, Q11
“M1: Substitute both limits, subtract and set the result equal to ln(7). A1: Reach a=ln(7/2) and state k=7/2 or an exact equivalent” — Pearson 9MA0, Sample Paper 1, Q4

What the examiner wants

The simplest correct strategy, kept exact. The June 2024 Paper 1 report says most found the area of the triangle and subtracted the area under the curve, but a significant proportion used Pythagoras and trigonometry for the triangle, which led to errors or premature rounding, and the Paper 2 report notes sign errors when integration by parts was applied twice.

Exam tip

For an enclosed area, solve for every intersection, identify upper minus lower, and check that the final area is positive.

Worked practice

Tier 1 · Easy

Find the area between the line

y=2x+1

, the

x

-axis, and the lines

x=0

and

x=3

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Area $=12$ square units	3
Notes
The line is above the axis on the interval, so the area is $\int_0^3(2x+1)\,\mathrm dx=[x^2+x]_0^3=9+3=12$ .

(3 marks)

Tier 2 · Standard

Find the finite area between the curve

y=x^2-4x+3

and the

x

-axis.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Area $=\dfrac43$ square units	4
Notes
The curve meets the $x$ -axis where $(x-1)(x-3)=0$ , so the limits are $1$ and $3$ . The quadratic is below the axis between these roots, so the area is $\int_1^3-(x^2-4x+3)\,dx=[-x^3/3+2x^2-3x]_1^3=0-(-4/3)=4/3$ square units.

(4 marks)

Tier 3 · Hard

A curve is given parametrically by

x=t^2

and

y=t^3

for

0\leq t\leq2

. Find the exact area between the curve, the

x

-axis and the line

x=4

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Area $=\frac{64}{5}$ square units	6
Notes
For a parametric curve, area is $\int y\,\mathrm dx=\int_0^2y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt$ . Here $\frac{\mathrm dx}{\mathrm dt}=2t$ , so the area is $\int_0^2t^3(2t)\,\mathrm dt=2\int_0^2t^4\,\mathrm dt=2[\frac15t^5]_0^2=\frac{64}{5}$ .

(6 marks)

Tier 1 · Easy

Find the exact area under the curve

y=4e^{-x}

between

x=0

and

x=\ln3

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	Area $=\dfrac83$ square units	3
Notes
The curve is positive, so the area is $\int_0^{\ln3}4e^{-x}\,\mathrm dx=[-4e^{-x}]_0^{\ln3}=-\frac43+4=\frac83$ square units.

(3 marks)

Tier 2 · Standard

Find the exact area between the curves

y=e^x

and

y=1+x

, and the lines

x=0

and

x=1

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Area $=e-\dfrac52$ square units	4
Notes
On $0\leq x\leq1$ , $e^x\geq1+x$ . Therefore the area is $\int_0^1(e^x-1-x)\,\mathrm dx=[e^x-x-\frac12x^2]_0^1=e-\frac52$ square units.

(4 marks)

Tier 3 · Hard

Find the exact total area between the curves

y=\sin x

and

y=\cos x

for

0\leq x\leq\dfrac{\pi}{2}

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Area $=2\sqrt2-2$ square units	5
Notes
The curves intersect when $\sin x=\cos x$ , so $x=\pi/4$ . The total area is $\int_0^{\pi/4}(\cos x-\sin x)\,\mathrm dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,\mathrm dx$ . Each integral equals $\sqrt2-1$ , giving $2\sqrt2-2$ square units.

(5 marks)

Tier 2 · Standard

Find the total area between the curve

y=x(x-1)(x-2)

and the

x

-axis for

0\leq x\leq2

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Area $=\dfrac12$ square units	5
Notes
The curve is above the axis on $(0,1)$ and below it on $(1,2)$ . An antiderivative of $x(x-1)(x-2)=x^3-3x^2+2x$ is $F(x)=x^4/4-x^3+x^2$ . Hence the total area is $[F(x)]_0^1-[F(x)]_1^2=1/4-(-1/4)=1/2$ square units.

(5 marks)

Tier 3 · Hard

The curve

y=kx-x^2

, where

k>0

, and the

x

-axis enclose a finite region of area

\frac92

square units. Find

k

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$k=3$	5
Notes
The curve meets the $x$ -axis where $x(k-x)=0$ , so the enclosed region lies between $x=0$ and $x=k$ . Its area is $\int_0^k(kx-x^2)\,\mathrm dx=[kx^2/2-x^3/3]_0^k=k^3/6$ . Thus $k^3/6=9/2$ , so $k^3=27$ . Since $k>0$ , $k=3$ .

(5 marks)

Tier 3 · Hard

The curve

y=4-x^2

and the

x

-axis enclose a finite region. The line

x=a

, where

-2<a<2

, cuts off an area of

9

square units between

x=-2

and

x=a

. Find

a

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$a=1$	6
Notes
The curve meets the axis at $x=-2$ and $x=2$ . The stated area condition gives $\int_{-2}^{a}(4-x^2)\,\mathrm dx=9$ , so $4a-a^3/3+16/3=9$ . Hence $a^3-12a+11=0$ , which factors as $(a-1)(a^2+a-11)=0$ . The other roots are $(-1\pm3\sqrt5)/2$ , both outside $(-2,2)$ , so the pinned value is $a=1$ .

(6 marks)

Q10

Tier 3 · Hard

10.

A curve is given parametrically by

x=2+t^2

and

y=t^3-4t

for

0\leq t\leq3

. Find the exact area enclosed by the curve and the chord joining its endpoints.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Area $=\dfrac{423}{10}$ square units	6
Notes
The endpoints are $(2,0)$ and $(11,15)$ , so the chord is $y=5(x-2)/3$ . At parameter $t$ the chord has height $5t^2/3$ . Also $5t^2/3-(t^3-4t)=t(3-t)(t+4/3)\geq0$ for $0\leq t\leq3$ , so the chord is above the curve. Since $\mathrm dx/\mathrm dt=2t$ , the area is $\int_0^3[5t^2/3-(t^3-4t)]2t\,\mathrm dt=\int_0^3(10t^3/3-2t^4+8t^2)\,\mathrm dt=423/10$ square units.

(6 marks)

Verified exam appearances

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