Integration with function as boundary I have to find the function F defined trough the integral: 
A) $\displaystyle F(x)= \int_0^{1-x} (1-2t+3t^2) \, dt $
B) $\displaystyle F(x)= \int_x^{x^2} (\sqrt t + \sqrt{t}^3) \, dt $
C) $\displaystyle F(x)= \int_0^{x^2} (1/2-\sin t) \, dt $
I tried solving the integrals:
A)  $\displaystyle F(y)= \int_0^{1-x} (1-2t+3t^2) \, dt = -x^3+2x^2-2x+1$
B) I do not know how to integrate this...
C) $ \displaystyle F(x)= \int_0^{x^2} (1/2-\sin t) \, dt = \frac{x^2}{2} +\cos(x^2)-1$
Is this correct ? Please if you can explain me how to solve it if I am wrong and why... Thanks for your time and help.
 A: Let's have a look at B) having in mind that an anti derivative of $t^{\alpha}$ is ${t^{\alpha+1}\over \alpha+1}$
$$\begin{align}\int_x^{x^2}\left(t^{1\over 2}+t^{3\over 2}\right)dt&=\left[{2t^{3\over 2}\over 3}\right]_x^{x^2}+\left[{2t^{5\over 2}\over 5}\right]_x^{x^2}\\
&={2\over 3}\left(x^3-\sqrt{x}^3\right)+{2\over 5}\left(x^5-\sqrt{x}^5\right)\end{align}$$
A: (A) looks correct. (B) is very straightforward once you realize $ \sqrt t $ = $t^ \frac{1}{2}$ and similarly, $ \sqrt (t^3) $ = $t^ \frac{3}{2}$.So now it's simply a matter of using the ordinary integration rules: 
$\displaystyle F(x)= \int_x^{x^2} (\sqrt t + \sqrt{t}^3) \, dt $ = 
$\displaystyle F(x)= \int_x^{x^2} (t^ \frac{1}{2} + t^ \frac{3}{2})\,dt = \frac{2}{3} t^ \frac{3}{2} + \frac{2}{5}  t^ \frac{5}{2}|_{x}^{x^2}=\frac{2}{3}(x^ 3) + \frac{2}{5} (x^ 5))- (\frac{2}{3} x^ \frac{3}{2} + \frac{2}{5}  x^ \frac{5}{2})$ 
You could rearrange (B) to collect like terms,but in this case I don't think it matters unless your prof is anal about layout. (C) also looks correct. 
