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)#