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 int_0^(x^2)sin(t^2)dt

Let's say u=x^2. We may now rewrite our integral:

We have int_0^usin(t^2)dt

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

d/dxint_0^xf(t)dt=f(x)

Therefore,

d/dxint_0^usin(t^2)dt=sin(u^2)(du)/dx

(We include (du)/dx because u=x^2 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.)

(du)/dx=2x

u^2=(x^2)^2=x^4

Rewriting in terms of x:

d/dxint_0^(x^2)sin(t^2)dt=2xsin(x^4)