Topic 7 of 7

Fundamental theorem of calculus

Step 15 · Calculus 1 topics

Link derivatives and integrals: differentiate an accumulated area, evaluate definite integrals with antiderivatives, and add up rates to find net change.

Accumulation functions

Let A(x) be the area under f from a fixed point a up to x. When x moves right by a sliver h, the area grows by about f(x) × h. So the rate at which area builds up is the height of the curve: A′(x) = f(x). This is the first part of the fundamental theorem.

\frac{d}{dx}\int_{a}^{x} f(t)\,dt = f(x)

A(4) is the shaded area under y = x ÷ 2 + 1 up to x = 4. Moving x a little further adds a thin strip about f(4) = 3 tall, so A grows at rate 3.

Example: Differentiating an integral

Find $\frac{d}{dx}\int_{1}^{x} \sqrt{t^{3} + 1}\,dt$ .

By the first part of the theorem, the answer is the integrand with x in place of t.

Answer $\sqrt{x^{3} + 1}$

Example: With the chain rule

Find $\frac{d}{dx}\int_{0}^{x^{2}} \cos t\,dt$ .

The upper limit is x², so $\cos(x^{2})$ is multiplied by the derivative of x², which is 2x.

Answer $2x\cos(x^{2})$

Evaluating definite integrals

The second part: if F is any antiderivative of f, the integral from a to b is F(b) − F(a). The + C cancels in the subtraction, so any antiderivative works.

\int_{a}^{b} f(x)\,dx = F(b) - F(a)

Example: A polynomial

Find $\int_{1}^{3} (3x^{2} - 2x)\,dx$ .

An antiderivative is x³ − x².
F(3) is 27 − 9, which is 18, and F(1) is 1 − 1, which is 0.

Answer18

Example: Trig

Find $\int_{0}^{\pi} \sin x\,dx$ .

An antiderivative is −cos x.
−cos π − (−cos 0) is 1 + 1.

Answer2

Example: Check against the rectangles

Find $\int_{0}^{2} x^{2}\,dx$ exactly.

$\frac{x^{3}}{3}$ from 0 to 2 gives $\frac{8}{3}$ .
The 4-rectangle estimate of 3.75 was too high, as expected.

Answer $\frac{8}{3}$

Net change

If a quantity Q changes at rate Q′, integrating the rate from a to b gives the total change Q(b) − Q(a). Rates in, amounts out.

Example: Displacement

A car's velocity is v(t) = 3t² − 12 m/s for 0 ≤ t ≤ 3. Find its displacement.

An antiderivative is t³ − 12t.
At 3 it is 27 − 36, which is −9; at 0 it is 0.

Answer−9 m: it ends 9 m behind where it started

Example: Distance traveled

For the same car, find the total distance traveled.

v is negative for t < 2 and positive after, so split there.
From 0 to 2, t³ − 12t goes from 0 to −16: 16 m backward.
From 2 to 3, it goes from −16 to −9: 7 m forward.

Answer23 m

Try it yourself

Work each one on paper first, then open the answer to check.

Find $\int_{0}^{2} (4x^{3} + 1)\,dx$ .
Show answer
18
Find $\int_{1}^{e} \frac{1}{x}\,dx$ .
Show answer
1
Find $\int_{0}^{\pi/2} \cos x\,dx$ .
Show answer
1
Find $\frac{d}{dx}\int_{2}^{x} t^{4}\,dt$ .
Show answer
x⁴
Water flows into a tank at 4t liters per minute. How much flows in from t = 0 to t = 5?
Show answer
50 liters
A particle's velocity is v(t) = t − 2 m/s. Find its displacement from t = 0 to t = 4.
Show answer
0 m
For the same particle, find the distance traveled from t = 0 to t = 4.
Show answer
4 m