The Fundamental Theorem of Calculus: evaluating integrals

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Calculus BC (AP-aligned) › Integrals and accumulation

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The Fundamental Theorem of Calculus: evaluating integrals

Warm-up

Three quick questions from earlier lessons.

Discover

$v(t)$ is a velocity and $s(t)$ the position, so $s' = v$ . The area under $v$ on $[a, b]$ is the change in position. Which expression is it?

$s(b) - s(a)$
$s(a) - s(b)$

Key idea

No more limits of sums: use an antiderivative.

Fundamental Theorem of Calculus: if

f

is continuous on

[a, b]

and

F' = f

, then

\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)

.

Worked example

Evaluate $\displaystyle\int_1^3 (x^2 + 1)\,dx$ .

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

An antiderivative.

F(3) - F(1) = 12 - \tfrac{4}{3}

F(3) = 12, F(1) = 4/3.

= \tfrac{32}{3}

Compute.

Your turn, guided

Your turn: $\displaystyle\int_0^2 3x^2\,dx$ .

= 8

F(x) = x³: F(2) − F(0).

Practice ★

Evaluate $\displaystyle\int_0^{ 5 } x^2\,dx$ .
Evaluate $\displaystyle\int_0^{ 9 } x^2\,dx$ .
Evaluate $\displaystyle\int_0^{ 8 } x^2\,dx$ .

Watch out!

Practice ★★

Evaluate $\displaystyle\int_0^{\pi/5} \cos x\,dx$ . Give an exact value.
Evaluate $\displaystyle\int_0^{\pi/3} \cos x\,dx$ . Give an exact value.

Evaluate $\displaystyle\int_{ -1 }^{ 4 } (2x + 9)\,dx$ .
Evaluate $\displaystyle\int_{ 1 }^{ 4 } (2x + 7)\,dx$ .

Boss problem. Find the area of the region between the parabola $y = 4 - x^2$ and the $x$ -axis.

Exit ticket

Three questions to check you have got it.

Evaluate $\displaystyle\int_{ -1 }^{ 4 } (2x + 9)\,dx$ .
Evaluate $\displaystyle\int_{ 1 }^{ 4 } (2x + 7)\,dx$ .
Evaluate $\displaystyle\int_{ -2 }^{ 5 } (2x + 8)\,dx$ .

Summary card

Fundamental Theorem (evaluation)

$\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$ , with $F' = f$ .

Watch out: upper minus lower; check the antiderivative's signs.

End of the lesson

Next lesson: The average value of a function Back to the chapter

Answers: The Fundamental Theorem of Calculus: evaluating integrals

A

$\frac{125}{3}$
$243$
$\frac{512}{3}$

B

$\sqrt{\frac{5}{8} - \frac{\sqrt{5}}{8}}$
$\frac{\sqrt{3}}{2}$

C

$60$
$36$

D

$\frac{32}{3}$

E

$60$
$36$
$77$

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