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)dtx20sin(t2)dt

Let's say u=x^2u=x2. We may now rewrite our integral:

We have int_0^usin(t^2)dtu0sin(t2)dt

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

d/dxint_0^xf(t)dt=f(x)ddxx0f(t)dt=f(x)

Therefore,

d/dxint_0^usin(t^2)dt=sin(u^2)(du)/dxddxu0sin(t2)dt=sin(u2)dudx

(We include (du)/dxdudx because u=x^2u=x2 and we're still differentiating with respect to xx. This is the Chain Rule combined with the first part of the Fundamental Theorem of Calculus.)

(du)/dx=2xdudx=2x

u^2=(x^2)^2=x^4u2=(x2)2=x4

Rewriting in terms of x:x:

d/dxint_0^(x^2)sin(t^2)dt=2xsin(x^4)ddxx20sin(t2)dt=2xsin(x4)