# Math Help - Definite Intergration

1. ## Definite Intergration

Hi
i few question i am having trouble finding the correct answer:
1) $\int_0^\frac{\pi}{2}5x+sin2x dx$

i got $\frac{5\pi^2}{8}+\frac{1}{2}$ however the answers says its $\frac{5\pi^2}{8}+1$

2) $\int_1^4(2+\frac{1}{x})^2 dx$

$\frac{(2+\frac{1}{x})^3}{3}$

$\frac{(2+\frac{1}{4})^3}{3}-\frac{(2+\frac{1}{1})^3}{3}$

which gives me $\frac{33}{4}$ and this is incorrect.
3) $\int_0^1x^2(1-x) dx$
someone tell me how would i start this off.

P.S

2. ## Rockhard

You can break the integral up separately as such

$\int_0^{\frac{\pi}{2}} 5x$

and

$\int_0^{\frac{\pi}{2}}sin(2X)$

you can calculate each definite integral then combine them using addition in the proper order the equation was in evaluating the first integral yields

$\int_0^{\frac{\pi}{2}} 5x = \frac{5x^2}{2}$

And the second, you can use a simple U-Substitution as such

$u = 2x$
$du = 2 dx$
$\frac{1}{2}du=dx$

Now integrating

$\frac{1}{2}\int_0^{\frac{\pi}{2}}sin(u) = -\frac{1}{2}cos(x)$

3. Originally Posted by Paymemoney

3) $\int_0^1x^2(1-x) dx$
someone tell me how would i start this off.
Expand the bracket
$\int_0^1 x^2-x^3 dx$

= $x^3/3-x^4/4$ |x=1, 0

4. Perhaps if you would review some basic algebraic simplification, it would be of some good benefit in improving your integration skills, and more in depth into the concept of substitution

5. Originally Posted by Paymemoney
2) $\int_1^4(2+\frac{1}{x})^2 dx$

$\frac{(2+\frac{1}{x})^3}{3}$

$\frac{(2+\frac{1}{4})^3}{3}-\frac{(2+\frac{1}{1})^3}{3}$

which gives me $\frac{33}{4}$ and this is incorrect.
$\int_1^4 (2+\frac{1}{x})^2 ~ dx = \int_1^4 4 + \frac{4}{x} + \frac{1}{x^2} ~ dx = \int_1^4 4 ~ dx + 4\int_1^4 \frac{1}{x} ~ dx + \int_1^4 \frac{1}{x^2} ~ dx$
You may only use the rule that you used if you have an integral of the form $\int x^n$ for $n\neq -1$.