Steply

d/dx ∫ from 0 to x^3 of sin(t) dt

Step by step, with the reason for every step · The Fundamental Theorem, part 1 · Calculus 1

Step by step

The question

Question: find the derivative of ∫ from 0 to x³ of sin t dt.

ddx∫ from 0 to x³ of sin t dt =

1 The rule

The rule: the derivative of the area from a to x is the function at x.

Just put x in place of t.

ddx∫ from a to x of f(t) dt = f(x)

Why?

Look at the area under f from 0 to x.

Move x a tiny bit to the right.

The area grows by a thin strip.

Its height is f(x).

So the area grows at the rate f(x).

2 Put the top in place of t

The top is not just x: it is x³.

Call it u.

Its slope is u′.

top: u = x3, u′ = 3x²

The chain rule: put u in place of t, then multiply by u′.

ddx∫ from 0 to x³ of sin t dt = f(u) × u′

Put x³ in place of t, times 3x².

= sin(x3) × 3x²

Careful!

Do not stop at f of u.

The top is x³, so we must multiply by its slope, 3x².

Careful: not sin(x3) ✗

Write the 3x² in front: 3x² sin(x³).

= 3x² sin(x3)

3 Check

Let's check at x = 1.

Our answer gives 2.5244, in radians.

answer at x = 1: 2.5244

Now the slope of the area function F between x = 1 and 1.001, with a calculator: about 2.529.

The same!

Right!

F(1.001) − F(1)0.001 ≈ 2.529

Let's look back.

First: the rule.

Then: put the top in place of t.

Last: check.

Answer
= 3x² sin(x3)

Your turn

Try this one yourself, then check:

ddx∫ from 1 to x³ of 5 sin t dt
Show the answer
3x² 5sin(x3)

More from The Fundamental Theorem, part 1

The whole lesson: The Fundamental Theorem, part 1 →