Tìm hàm số f(x), biết rằng: a) f'(x) = 2x^3 – 4x + 1, f(1) = 0; b) f'(x) = 5cosx – sinx, f ( π 2 ) = 1 .

a) \[f\left( x \right) = \int {f'\left( x \right)dx} \]

            \[ = \int {\left( {2{x^3} - 4x + 1} \right)dx} \]

            = \[\frac{{{x^4}}}{2} - 2{x^2} + x + C.\]

Mà f(1) = 0 ⇒ \[\frac{1}{2} - 2 + 1 + C = 0\]⇒ \[C = \frac{1}{2}\].

Vậy \[f\left( x \right) = \frac{{{x^4}}}{2} - 2{x^2} + x + \frac{1}{2}.\]

b) Ta có: \[f\left( x \right) = \int {f'\left( x \right)dx} \]

                        \[ = \int {\left( {5\cos x - \sin {\rm{x}}} \right)dx} \]

                        \[ = 5\sin x + \cos x + C\].

Mà \[f\left( {\frac{\pi }{2}} \right) = 1\] nên \[5\sin \left( {\frac{\pi }{2}} \right) + \cos \left( {\frac{\pi }{2}} \right) + C = 1\] hay \[5 + C = 1\] suy ra  C = −4.

Vậy f(x) = 5sinx + cosx – 4.