What is 8.3: Definite integrals and areas?

Evaluate _a^b f(x)\, dx by finding an antiderivative F and calculating F(b)-F(a). A definite integral is signed area.

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.

Explanation

Evaluate $\int_a^b f(x)\,\mathrm dx$ by finding an antiderivative $F$ and calculating $F(b)-F(a)$ . A definite integral is signed area.
Split at roots or intersections if the required geometric area includes regions below the axis or where the upper curve changes. For area between curves, solve their intersection equations first and integrate $y_{\text{upper}}-y_{\text{lower}}$ over each relevant interval.
A common error is to reverse the subtraction or to quote a negative definite integral as a negative geometric area.
Intersections satisfy $4x-x^2=x$ , so $x(3-x)=0$ and $x=0,3$ .
For a parametric curve, its signed area is $\int y\,\dfrac{dx}{dt}\,dt$ over the parameter interval, with limits ordered to match the traced direction.

For a parametric curve, vertical strips give area

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

Worked example

Find the finite area enclosed by the curve $y=4x-x^2$ and the line $y=x$ .

1.Intersections satisfy $4x-x^2=x$ , so $x(3-x)=0$ and $x=0,3$ .
2.The parabola is above the line between them.
3.Thus the area is $\int_0^3(3x-x^2)\,\mathrm dx=[\frac32x^2-\frac13x^3]_0^3=\frac{27}{2}-9=\frac92$ .

Answer: Area $=\frac92$ square units

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.

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

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

Other points in 8 Integration

Want help turning this into marks?

Bring 8.3 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call