A concept question about integration: int[0,2] [ 3 f '(x) - 4x g '(x^2) ] dx

[hades94720]

I have this worked solution:

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(a) Evaluate the definite integral.

. . . . .\(\displaystyle \mbox{(i) }\, \)\(\displaystyle \displaystyle \int_0^2\, \bigg[\, 3\, f'(x)\, -\, 4x\, g'(x^2)\, \bigg]\, dx\)

Solution:

. . . . . . . . .\(\displaystyle \displaystyle \int_0^2\, \bigg[\, 3\, f'(x)\, -\, 4x\, g'(x^2)\, \bigg]\, dx\, =\, \bigg[\, 3\, f(x)\, -\, 2\, g(x^2)\, \bigg]\, \bigg|_0^2\)

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I don't understand why the '4x' would be changed to 2; it should be 2x^2 (by 4* x^(1+1)/(1+1)),

isn't it ?

Should it be [3f(x)- 2x^2 g(x^2)] ?

Thanks.

 

[Deleted member 4993]

I don't understand why the '4x' would be changed to 2; it should be 2x^2 (by 4* x^(1+1)/(1+1)),

isn't it ?

Should it be [3f(x)- 2x^2 g(x^2)] ?

Thanks.

\(\displaystyle \displaystyle{\dfrac{d}{dx}\left [g(x^2)\right ] \ = \ g'(x^2) * \left [\dfrac{d}{dx}(x^2) \right ]\ = \ g'(x^2) * (2x) }\).......... chain rule