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

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Answer 1 · 2018-04-24T05:52:56

Please refer to the Discussion in Explanation.

Suppose that, #intsint^2dt=f(t)+C........(square)#.

Therefore, by Definition of Integral, #f'(t)=sint^2......(square^1)#.

From #(square)#, on using the Fundamental Theorem of Calculus,

#int_(x^2)^0sint^2dt=[f(t)]_(x^2)^0=f(0)-f(x^2)#.

#:. d/dx{int_(x^2)^0sint^2dt}=d/dx{f(0)-f(x^2)}#,

#=0-d/dx{f(x^2)}...........[because," f(0) is constant]"#,

#=-f'(x^2)*d/dx{x^2}........[because," the Chain Rule]"#,

#=(-f'(x^2))(2x)#,

#=-2xf'(x^2)#,

#=-2xsin(x^4)..........[because, (square^1)]#.

We conclude that the statement is false.