Okay, I'm not entirely sure I did these correctly. Any help would be greatly appreciated!
1. [1, 2] (3/x^2 - 1) dx
2. [0, pi] (1+sinx) dx
3. [0,3] |2x-3| dx
Thanks!
1. Power rule and evaluate.
2. Split the integral into two ones.
3. This is
$\displaystyle \int_0^3 {\left| {2x - 3} \right|\,dx} = \int_0^{3/2} {\left| {2x - 3} \right|\,dx} + \int_{3/2}^3 {\left| {2x - 3} \right|\,dx} .$
In the first piece, for $\displaystyle 0 \le x \le \frac{3}
{2} \implies \left| {2x - 3} \right| = 3 - 2x.$
In the second piece, for $\displaystyle \frac{3}
{2} \le x \le 3 \implies \left| {2x - 3} \right| = 2x-3.$
Hence
$\displaystyle \int_0^3 {\left| {2x - 3} \right|\,dx} = \int_0^{3/2} {(3-2x)\,dx} + \int_{3/2}^3 {(2x - 3)\,dx} .$
You should be able to do that.
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Lookin' at you first question it could be $\displaystyle \frac{3}
{{x^2 - 1}}.$ In that case, split the original ratio into two fractions.
Look at the definition of the absolute value function:
$\displaystyle |x|=\left\{\begin{array}{rr}x,\text{ if }x \geq 0\\
-x,\text{ if }x<0\end{array}\right.$
So
$\displaystyle |2x - 3|=\left\{\begin{array}{rr}2x - 3,\text{ if }2x - 3 \geq 0\\
-(2x - 3),\text{ if }2x - 3<0\end{array}\right.$
-Dan