Step by step
The question
Question: find the derivative of ∫ from 0 to x of 1/(t² + 1) dt.
ddx∫ from 0 to x of 1t2 + 1 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)
2 Put x in place of t
Put x in place of t.
So the derivative is 1/(x² + 1).
ddx∫ from 0 to x of 1t2 + 1 dt = 1x2 + 1
Careful!
There is nothing to take away.
The derivative of the area is just f at x.
The bottom number 0 does not matter.
Careful: not 1t2 + 1 − f(0) ✗
3 Check
Let's check at x = 1.
Our answer gives 0.5.
answer at x = 1: 0.5
Now the slope of the area function F between x = 1 and 1.001, with a calculator: about 0.5.
The same!
Right!
F(1.001) − F(1)0.001 ≈ 0.5
Let's look back.
First: the rule.
Then: put x in place of t.
Last: check.
Answer
Careful: not 1t2 + 1 − f(0) ✗