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Thread: Integrals

  1. #1
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    Integrals

    Evaluate the indefinite and definite integrals.

    1) from -2 to 3 (25)^2x+1 dx

    2) 7x6^1-2x^2 dx

    These are two types of problems I am having trouble with. Please help.
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  2. #2
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    Hello, bigton!

    There is a formula for these exponential integrals:

    . . $\displaystyle \int b^u\,du \;=\;\frac{b^u}{\ln(b)}+ C$



    $\displaystyle 1)\;\;\int^3_{\text{-}2}(25)^{2x+1} dx$

    Let: $\displaystyle u \:=\:2x+1\quad\Rightarrow\quad du \:=\:2\,dx \quad\Rightarrow\quad dx \:=\:\tfrac{1}{2}du $

    Substitute: .$\displaystyle \int25^u\cdot\tfrac{1}{2}du \;=\;\tfrac{1}{2}\int 25^u\,du \;=\;\frac{1}{2}\,\frac{25^u}{\ln(25)} + C$

    Back-substitute: .$\displaystyle \frac{25^{2x+1}}{2\ln(25)}\,\bigg]^3_{\text{-}2} \;=\;\frac{25^7}{2\ln(25)} - \frac{25^{-1}}{2\ln(25)} \quad\hdots\;\text{ etc.}$




    $\displaystyle 2)\;\;\int 7x\!\cdot\!6^{1-2x^2}\,dx$

    We have: .$\displaystyle 7\!\int 6^{1-2x^2}(x\,dx)$

    Let: $\displaystyle u \:=\:1-2x^2\quad\Rightarrow\quad du \:=\:\text{-}4x\,dx \quad\Rightarrow\quad x\,dx \:=\:\text{-}\tfrac{1}{4}du$

    Substitute: .$\displaystyle 7\!\int 6^u(\text{-}\tfrac{1}{4}du) \;=\;-\tfrac{7}{4}\!\int6^u\,du \;=\;-\frac{7}{4}\,\frac{6^u}{\ln(6)} + C$

    Back-substitute: .$\displaystyle -\frac{7\cdot6^{1-2x^2}}{4\cdot\ln(6)} + C $

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  3. #3
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    What does the etc... mean
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