calculus help?

Question

Consider the function f(x) whose second derivative is f ''(x) = 4 x + 4 \sin (x). If f(0) = 2 and f '(0) = 4, what is f(x)?

Answer 1

f''(x) = 4x + 4 sin x
f'(x) = 2x² - 4 cos x + C, where C is a constant.
f'(0) = - 4 + C = 4, so C=8
f'(x) = 2x² - 4 cos x + 8
f(x) = 2x³/3 - 4 sin x + 8x + K, where K is a constant
f(0) = K = 2

Thus f(x) = 2x³/3 - 4 sin x + 8x + 2. And we are done.

Answer 2

f '(x) = 2x^2 + integ(4\sinx)+C
Find C by setting 4 = 2x^2 + integ(4\sinx) with x = 0
Now integrate again and find new C 2 = result with x = 0

Sorry, I did not integrate (4\sinx) for you because I do not understand the \ symbol.

But the above should show you how to do it.