D/dx[∫sin(t²)dt from x² to 0]=-sin(x²).This statement is true or false?Please give reasons for your answers.

1 Answer
Feb 25, 2018

False.

Explanation:

False.

We have x20sin(t2)dt

Let's say u=x2. We may now rewrite our integral:

We have u0sin(t2)dt

The first part of the Fundamental Theorem of Calculus tells us that

ddxx0f(t)dt=f(x)

Therefore,

ddxu0sin(t2)dt=sin(u2)dudx

(We include dudx because u=x2 and we're still differentiating with respect to x. This is the Chain Rule combined with the first part of the Fundamental Theorem of Calculus.)

dudx=2x

u2=(x2)2=x4

Rewriting in terms of x:

ddxx20sin(t2)dt=2xsin(x4)