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Thread: integrals help

  1. #1
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    integrals help

    Compute an antiderivative for each of the following functions:

    $\displaystyle a) f(x) = \frac{2x^3+1}{x^2}$
    $\displaystyle b) g(x) = \sqrt{x}+x^\frac{7}{3}$
    $\displaystyle c) h(x) = csc(2x)cot(2x)$

    I was never taught integrals properly, only the basics where u change $\displaystyle x^2$ to $\displaystyle x^3/3$.


    can anyone help? thanks!
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  2. #2
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    Quote Originally Posted by break View Post
    Compute an antiderivative for each of the following functions:

    $\displaystyle a) f(x) = \frac{2x^3+1}{x^2}$
    $\displaystyle b) g(x) = \sqrt{x}+x^\frac{7}{3}$
    $\displaystyle c) h(x) = csc(2x)cot(2x)$

    I was never taught integrals properly, only the basics where u change $\displaystyle x^2$ to $\displaystyle x^3/3$.


    can anyone help? thanks!


    For a polynomial $\displaystyle ax^n$ the integral is given by $\displaystyle a\int x^ndx=\frac{a}{n+1}x^{n+1}+C$. Convince yourself this is true by differentiating.

    The first one can be written as a sum:

    $\displaystyle \int\frac{2x^3+1}{x^2}dx=\int2xdx+\int\frac{1}{x^2 }dx$

    Use the formula. Remember that $\displaystyle \frac{1}{x^2}=x^{-2}$.

    Problem (b) just requires the formula. Remember that $\displaystyle \sqrt{x}=x^{\frac{1}{2}}$.

    For (c) rememeber that $\displaystyle \frac{d}{du}csc(u)=-csc(u)cot(u)$

    So if you let $\displaystyle u=2x$ so that $\displaystyle du=2dx$ then $\displaystyle csc(2x)cot(2x)dx=\frac{1}{2}csc(u)cot(u)du$. Does this make sense? Do you see what the integral is?
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  3. #3
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    I had to edit my original post for latex error. Sorry about that.
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  4. #4
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    Quote Originally Posted by adkinsjr View Post
    The first one can be written as a sum:

    $\displaystyle \int\frac{2x^3+1}{x^2}dx=\int2xdx+\int\frac{1}{x^2 }dx$

    Use the formula. Remember that $\displaystyle \frac{1}{x^2}=x^{-2}$.
    ok so this turns to

    $\displaystyle \int\frac{2x^2}{2} + \int\frac{x^-1}{-1}$

    $\displaystyle \int x^2dx + \int\frac{-1}{x}dx$

    i think i did something wrong
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  5. #5
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    Quote Originally Posted by break View Post
    Compute an antiderivative for each of the following functions:

    $\displaystyle a) f(x) = \frac{2x^3+1}{x^2}$
    $\displaystyle b) g(x) = \sqrt{x}+x^\frac{7}{3}$
    $\displaystyle c) h(x) = csc(2x)cot(2x)$

    I was never taught integrals properly, only the basics where u change $\displaystyle x^2$ to $\displaystyle x^3/3$.


    can anyone help? thanks!
    Alternatively, for c), remember

    $\displaystyle \csc{(2x)}\cot{(2x)} = \frac{1}{\sin{(2x)}}\cdot \frac{\cos{(2x)}}{\sin{(2x)}}$

    $\displaystyle = \frac{\cos{(2x)}}{\sin^2{(2x)}}$.


    Now use a $\displaystyle u$ substitution.

    Let $\displaystyle u = \sin{(2x)}$ so that $\displaystyle \frac{du}{dx} = 2\cos{(2x)}$.


    Then you have

    $\displaystyle \int{\csc{(2x)}\cot{(2x)}\,dx} = \int{\frac{\cos{(2x)}}{\sin^2{(2x)}}\,dx}$

    $\displaystyle = \frac{1}{2}\int{\frac{1}{\sin^2{(2x)}}[2\cos{(2x)}]\,dx}$

    $\displaystyle = \frac{1}{2}\int{\frac{1}{u^2}\,\frac{du}{dx}\,dx}$

    $\displaystyle = \frac{1}{2}\int{u^{-2}\,du}$

    $\displaystyle = -\frac{1}{2}u^{-1} + C$

    $\displaystyle = -\frac{1}{2}\cdot\frac{1}{\sin{(2x)}} + C$

    $\displaystyle = -\frac{1}{2}\csc{(2x)} + C$.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Then you have

    $\displaystyle \int{\csc{(2x)}\cot{(2x)}\,dx} = \int{\frac{\cos{(2x)}}{\sin^2{(2x)}}\,dx}$

    $\displaystyle = -\frac{1}{2}\int{\frac{1}{\sin^2{(2x)}}[-2\cos{(2x)}]\,dx}$
    i got lost here. where did u get the -1/2 from? is there a rule for this?
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  7. #7
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    Quote Originally Posted by break View Post
    ok so this turns to

    $\displaystyle \int\frac{2x^2}{2} + \int\frac{x^-1}{-1}$

    $\displaystyle \int x^2dx + \int\frac{-1}{x}dx$

    i think i did something wrong
    You're way off. The integral $\displaystyle \int 2xdx=\frac{2}{3}x^3+C$ is given by the formula $\displaystyle \int ax^ndx=\frac{a}{n+1}x^{n+1}+C$. Do you see how I applied the formula? It works the same for all polynomials. See if you can do the same for the second term.

    Here is a tutorial that might help you with some of these. It seems like you don't have the basic concept of what an integral or anti-derivative is.

    Pauls Online Notes : Calculus I - Indefinite Integrals
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    Quote Originally Posted by break View Post
    i got lost here. where did u get the -1/2 from? is there a rule for this?
    1. Notice my edit.

    2. We need to change the integrand so that it has $\displaystyle \frac{du}{dx}$ in it.

    So we change $\displaystyle \cos{(2x)}$ into $\displaystyle \frac{1}{2}\cdot 2\cos{(2x)}$.

    Then I took the $\displaystyle \frac{1}{2}$ outside the integral.
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  9. #9
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    Quote Originally Posted by adkinsjr View Post
    You're way off. The integral $\displaystyle \int 2xdx=\frac{2}{3}x^3+C$ is given by the formula $\displaystyle \int ax^ndx=\frac{a}{n+1}x^{n+1}+C$. Do you see how I applied the formula? It works the same for all polynomials. See if you can do the same for the second term.

    Here is a tutorial that might help you with some of these. It seems like you don't have the basic concept of what an integral or anti-derivative is.

    Pauls Online Notes : Calculus I - Indefinite Integrals
    why does x skip over $\displaystyle x^2$ and go to $\displaystyle x^3$? I see how the formula says n+1 but, 1+1 = 3? i don't get it.
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  10. #10
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    Quote Originally Posted by break View Post
    why does x skip over $\displaystyle x^2$ and go to $\displaystyle x^3$? I see how the formula says n+1 but, 1+1 = 3? i don't get it.
    You're right

    It should be $\displaystyle x^2+C$.

    $\displaystyle \int 2xdx =\frac{2}{1+1}x^{1+1}+C=x^2+C$

    Sorry about that. That was just a stupid mistake on my part.
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  11. #11
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    Quote Originally Posted by adkinsjr View Post
    You're right

    It should be $\displaystyle x^2+C$.

    $\displaystyle \int 2xdx =\frac{2}{1+1}x^{1+1}+C=x^2+C$

    Sorry about that. That was just a stupid mistake on my part.
    haha my brain was seriously about to explode right now. phew.

    Quote Originally Posted by break View Post
    $\displaystyle \int\frac{2x^2}{2} + \int\frac{x^-1}{-1}$

    $\displaystyle \int x^2dx + \int\frac{-1}{x}dx$

    i think i did something wrong
    so i did get the first part right... right? haha. do i still have to plug in a + C early on?
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