What is 8.1: Fundamental theorem of calculus?

Integration is the reverse process of differentiation: if dy/ dx=f(x), then y= f(x)\, dx recovers the family of curves with that gradient function.

How many marks are 8.1 questions worth?

8.1 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has 8.1 appeared in?

We have not yet indexed a verified past-paper appearance for 8.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

8.1

Know and use the Fundamental Theorem of Calculus.

Draft — not yet indexed

Fundamental theorem of calculus

Worked answers and methods for 8.1 on Edexcel A-level Maths 9MA0.

Explanation

Integration is the reverse process of differentiation: if $\frac{\mathrm dy}{\mathrm dx}=f(x)$ , then $y=\int f(x)\,\mathrm dx$ recovers the family of curves with that gradient function.
Every indefinite integral needs a constant of integration: $\int f(x)\,\mathrm dx=F(x)+c$ .
One known point on the curve fixes the value of $c$ .
A definite integral is evaluated from any antiderivative: $\int_a^b f(x)\,\mathrm dx=F(b)-F(a)$ ; the constant of integration cancels in the subtraction.
A common error is to omit $+c$ from an indefinite integral, or to substitute the known point into $f(x)$ instead of the integrated function when finding $c$ .

Worked example

The function $f$ satisfies $f'(x)=3\sqrt{x}-\dfrac{2}{x^2}$ for $x>0$ , and $f(4)=10$ . Find $f(x)$ .

1.Rewrite as $f'(x)=3x^{1/2}-2x^{-2}$ and integrate: $f(x)=2x^{3/2}+2x^{-1}+c$ .
2.Substituting $x=4$ : $2(8)+\frac{2}{4}+c=10$ , so $16.5+c=10$ and $c=-\frac{13}{2}$ .
3.Hence $f(x)=2x^{3/2}+\frac{2}{x}-\frac{13}{2}$ .

Answer: $f(x)=2x^{3/2}+\dfrac{2}{x}-\dfrac{13}{2}$

Common mistakes

Don't use the supplied boundary value itself as the integration constant without substituting into the antiderivative.
Don't integrate the derivative but omit the arbitrary constant before using the given function value.

Exam tip

When recovering a function from its derivative, include the integration constant and determine it from the boundary condition.

Worked practice

Tier 1 · Easy

A curve has gradient function

\frac{\mathrm dy}{\mathrm dx}=6x^2-4

and passes through the point

(1,5)

. Find the equation of the curve.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$y=2x^3-4x+7$	3
Notes
Integrating the gradient function gives $y=2x^3-4x+c$ . Substituting $(1,5)$ : $5=2-4+c$ , so $c=7$ . Hence $y=2x^3-4x+7$ .

(3 marks)

Tier 2 · Standard

Given that

F(x)=\dfrac{x^4}{2}-x^3+4x

is an antiderivative of

2x^3-3x^2+4

, evaluate

\displaystyle\int_1^3(2x^3-3x^2+4)\,\mathrm dx

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$22$	4
Notes
Use the definite-integral result $\int_a^b f(x)\,dx=F(b)-F(a)$ . Here $F(3)=81/2-27+12=51/2$ and $F(1)=1/2-1+4=7/2$ . Therefore the integral is $51/2-7/2=22$ .

(4 marks)

Tier 3 · Hard

A curve

y=f(x)

satisfies

f''(x)=6x+2

. The gradient of the curve at

x=1

3

, and the curve passes through

(2,4)

. Find

f(x)

, and verify your answer by evaluating

\int_1^2 f'(x)\,\mathrm dx

and comparing it with

f(2)-f(1)

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$f(x)=x^3+x^2-2x-4$ $\int_1^2 f'(x)\,\mathrm dx=8=f(2)-f(1)$	5
Notes
Integrate once: $f'(x)=3x^2+2x+c_1$ ; from $f'(1)=3$ , $3+2+c_1=3$ so $c_1=-2$ . Integrate again: $f(x)=x^3+x^2-2x+c_2$ ; from $f(2)=4$ , $8+4-4+c_2=4$ so $c_2=-4$ . Then $f(2)-f(1)=4-(-4)=8$ , and $\int_1^2(3x^2+2x-2)\,\mathrm dx=\left[x^3+x^2-2x\right]_1^2=8-0=8$ , confirming $\int_a^b f'(x)\,\mathrm dx=f(b)-f(a)$ .

(5 marks)

Tier 1 · Easy

The function

H

is an antiderivative of

q

. Given that

\displaystyle\int_1^4q(x)\,\mathrm dx=9

and

H(1)=5

, find

H(4)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$H(4)=14$	2
Notes
By the Fundamental Theorem of Calculus, $\int_1^4q(x)\,\mathrm dx=H(4)-H(1)$ . Hence $9=H(4)-5$ , giving $H(4)=14$ .

(2 marks)

Tier 2 · Standard

A function

F

satisfies

F'(x)=3x^2+k

, where

k

is a constant. Given that

F(0)=2

and

F(2)=16

, find

k

and hence find

F(x)

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$k=3$ $F(x)=x^3+3x+2$	4
Notes
Integrating gives $F(x)=x^3+kx+C$ . The condition $F(0)=2$ gives $C=2$ . Then $F(2)=16$ gives $8+2k+2=16$ , so $k=3$ . Therefore $F(x)=x^3+3x+2$ .

(4 marks)

Tier 3 · Hard

A function

h

satisfies

h'(x)=|2x-4|

and

h(1)=7

. Find

h(5)

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$h(5)=17$	5
Notes
The expression inside the modulus changes sign at $x=2$ . Therefore $h(5)-h(1)=\int_1^2(4-2x)\,\mathrm dx+\int_2^5(2x-4)\,\mathrm dx=1+9=10$ . Hence $h(5)=7+10=17$ .

(5 marks)

Tier 2 · Standard

Let

F

be an antiderivative of

f

. Given that

\int_{-2}^{1}f(x)\,\mathrm dx=7

\int_{1}^{5}f(x)\,\mathrm dx=-4

and

F(5)=9

, find

F(-2)

. Hence evaluate

\int_{5}^{-2}f(x)\,\mathrm dx

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$F(-2)=6$ $\displaystyle\int_{5}^{-2}f(x)\,\mathrm dx=-3$	4
Notes
Add the adjacent integrals: $\int_{-2}^{5}f(x)\,\mathrm dx=7-4=3$ . By the Fundamental Theorem of Calculus, this equals $F(5)-F(-2)$ , so $3=9-F(-2)$ and $F(-2)=6$ . Reversing the limits changes the sign, hence $\int_{5}^{-2}f(x)\,\mathrm dx=-3$ .

(4 marks)

Tier 3 · Hard

A differentiable function

g

satisfies

g'(x)=(x-1)(x-2)

for

0\leq x\leq3

, and

g(0)=0

. Use the Fundamental Theorem of Calculus to find

g(1)

g(2)

and

g(3)

. Hence state the absolute maximum and minimum values of

g

on this interval.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$g(1)=\dfrac56$ , $g(2)=\dfrac23$ , $g(3)=\dfrac32$ Absolute maximum $\dfrac32$ at $x=3$ ; absolute minimum $0$ at $x=0$	6
Notes
An antiderivative of $g'(x)=x^2-3x+2$ is $G(x)=x^3/3-3x^2/2+2x$ . Since $g(a)-g(0)=\int_0^a g'(x)\,\mathrm dx=G(a)-G(0)$ and $g(0)=0$ , substitution gives $g(1)=5/6$ , $g(2)=2/3$ and $g(3)=3/2$ . The derivative is positive on $(0,1)$ , negative on $(1,2)$ and positive on $(2,3)$ . Comparing the endpoint and stationary values gives the stated absolute extrema.

(6 marks)

Tier 3 · Hard

A function

F

satisfies

F'(x)=3x^2-12x+9

and

F(0)=5

. Find all values of

x

0\leq x\leq3

for which

F(x)=5

. Hence find the exact area enclosed by the curve

y=F(x)

and the line

y=5

between consecutive points of intersection.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$F(x)=x^3-6x^2+9x+5$ ; the intersections are at $x=0$ and $x=3$ Area $=\dfrac{27}{4}$ square units	6
Notes
Integrating and using $F(0)=5$ gives $F(x)=x^3-6x^2+9x+5=x(x-3)^2+5$ . Therefore $F(x)=5$ when $x(x-3)^2=0$ , giving $x=0$ and $x=3$ in the stated interval. Since $x(x-3)^2\geq0$ on $[0,3]$ , the curve lies above the line there. Expanding the vertical difference gives the area $\int_0^3(x^3-6x^2+9x)\,\mathrm dx=81/4-54+81/2=27/4$ square units.

(6 marks)

Q10

Tier 3 · Hard

10.

A differentiable function

F

satisfies

F'(x)=ax+b

, where

a

and

b

are constants. Given that

F(2)-F(0)=6

F(5)-F(2)=3

and

F(0)=0

, use the Fundamental Theorem of Calculus to find

a

and

b

. Hence find

F(x)

and the absolute maximum value of

F

0\leq x\leq5

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$a=-\dfrac45$ , $b=\dfrac{19}{5}$ $F(x)=-\dfrac25x^2+\dfrac{19}{5}x$ Absolute maximum $\dfrac{361}{40}$ at $x=\dfrac{19}{4}$	7
Notes
By the Fundamental Theorem of Calculus, $F(2)-F(0)=\int_0^2(ax+b)\,\mathrm dx=2a+2b$ , so $a+b=3$ . Also $F(5)-F(2)=\int_2^5(ax+b)\,\mathrm dx=21a/2+3b=3$ . Solving gives $a=-4/5$ and $b=19/5$ . Integration and $F(0)=0$ then give $F(x)=-2x^2/5+19x/5$ . The only interior stationary point is $x=19/4$ ; the derivative changes from positive to negative there. Comparing with $F(0)=0$ and $F(5)=9$ gives the absolute maximum $F(19/4)=361/40$ .

(7 marks)

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