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Math Help - Definite integrals

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by alreadyinuse View Post
    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

    \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 0 \le x \le \frac{3}<br />
{2} \implies \left| {2x - 3} \right| = 3 - 2x.

    In the second piece, for \frac{3}<br />
{2} \le x \le 3 \implies \left| {2x - 3} \right| = 2x-3.

    Hence

    \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.

    ---

    Lookin' at you first question it could be \frac{3}<br />
{{x^2  - 1}}. In that case, split the original ratio into two fractions.
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  3. #3
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    But you're supposed to integrate both equations, correct?

    Thanks!
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  4. #4
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    For the third problem, the two remaining integrals?, of course.
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  5. #5
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    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?
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  6. #6
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    Quote Originally Posted by Destiny666 View Post
    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:
    |x|=\left\{\begin{array}{rr}x,\text{ if }x \geq 0\\<br />
-x,\text{ if }x<0\end{array}\right.

    So
    |2x - 3|=\left\{\begin{array}{rr}2x - 3,\text{ if }2x - 3 \geq 0\\<br />
-(2x - 3),\text{ if }2x - 3<0\end{array}\right.

    -Dan
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