Question

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

Answer 1

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)`