1. ## Definite integrals

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. [1, 2] (3/x^2 - 1) dx

2. [0, pi] (1+sinx) dx

3. [0,3] |2x-3| dx
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.

3. But you're supposed to integrate both equations, correct?

Thanks!

4. For the third problem, the two remaining integrals?, of course.

5. in your post above you split it inot 2 equations which makes sense and i can do but the equals part im haveing trouble understanding how does the abs of 2x-3 become 3-2x?

6. Originally Posted by Destiny666
in your post above you split it inot 2 equations which makes sense and i can do but the equals part im haveing trouble understanding how does the abs of 2x-3 become 3-2x?
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