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Math Help - Fundamental Theorem of Calculus

  1. #1
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    Fundamental Theorem of Calculus

    All my answers for the following questions are wrong, so please point out where I made the mistakes and how I could fix them or correct them for me! Thank you.

    Evaluate the given definite integral using the fundamental theorem of calculus.

    1) \int_{-1}^1 (2u^{\frac {1}{3}} - u^{\frac {2}{3}} \, du

    = \int_{-1}^1 \frac{2}{4} u^{\frac{1}{4}} - \frac{3}{5} u^{\frac{5}{3}} |_{-1}^1

    = - \frac{1}{10} - \frac{1}{10}

    = - \frac {1}{5}

    2) \int_{1}^2 (2x - 4)^4 \, dx

    = \frac{1}{5} (2x - 4)^5 \, |_1^2

    = \frac{1}{5} (4 - 4)^5 - \frac{1}{5} (2 - 4)^5

    = 6.4

    3) \int_0^4 \frac{1}{\sqrt{6t +1}} \, dt

    = \int_0^4 (6t + 1)^{- \frac{1}{2}} \, dt

    = 2(6t + 1)^{\frac{1}{2}} |_0^4

    = 8

    4) \int_0^1 (x^3 + x) \sqrt{x^4 + 2x^2 + 1} \, dx

    u = x^4 + 2x^2 + 1

    du = 4x^3 + 4x \, dx

    \int_0^1 (x^3 + x)u \frac{1}{4(x^3 + x)} \, du

    = \frac{1}{8}u|_1^4

    = \frac{15}{8}

    5) \int_2^{e + 1} \frac{x}{x - 1} \, dx

    u = x

    du = dx

    dv = (x - 1)^{-1} \, dx

    v = x - 1

    = x(x - 1)|_2^{e+1} + \int_2^{e + 1} x \, dx

    = x^2 - x + \frac{1}{2}x^2 |_2^{e + 1}

    = \frac{3}{2}(e + 1)^2 - (e + 1) + 3

    Textbook answers:

    1) -\frac{6}{5}

    2) 3.2

    3)  \frac{4}{3}

    4) \frac{7}{6}

    5) e

    P.S. How do you make that line "|" longer?
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Integration

    Hello Macleef

    Thanks for showing us your working - this must have taken you quite a long time!

    I've corrected the first mistakes you've made in each case - I hope you can finish them now.

    Grandad
    Quote Originally Posted by Macleef View Post
    All my answers for the following questions are wrong, so please point out where I made the mistakes and how I could fix them or correct them for me! Thank you.

    Evaluate the given definite integral using the fundamental theorem of calculus.

    1) \int_{-1}^1 (2u^{\frac {1}{3}} - u^{\frac {2}{3}} \, du

    = \int_{-1}^1 \frac{2}{4} \color{red}u^{\frac{1}{4}}\color{black} - \frac{3}{5} u^{\frac{5}{3}} |_{-1}^1

    \color{red}\int u^{\frac{1}{3}}du = \frac{3}{4}u^{\frac{4}{3}}

    ...

    2) \int_{1}^2 (\color{red}2\color{black}x - 4)^4 \, dx

    = \frac{1}{5}\color{red}\times\frac{1}{2}\color{blac  k} (2x - 4)^5 \, |_1^2

    ...

    3) \int_0^4 \frac{1}{\sqrt{6t +1}} \, dt

    = \int_0^4 (\color{red}6\color{black}t + 1)^{- \frac{1}{2}} \, dt

    = 2\color{red}\times\frac{1}{6}\color{black}(6t + 1)^{\frac{1}{2}} |_0^4

    ...

    4) \int_0^1 (x^3 + x) \sqrt{x^4 + 2x^2 + 1} \, dx

    u = x^4 + 2x^2 + 1

    du = 4x^3 + 4x \, dx

    \int_0^1 (x^3 + x)\color{red}\sqrt{u}\color{black} \frac{1}{4(x^3 + x)} \, du

    ...

    5) \int_2^{e + 1} \frac{x}{x - 1} \, dx

    \color{red}= \int_2^{e+1}\left(1+\frac{1}{x-1}\right)dx\color{black}

    \color{red}= \Big[x + \text{ln}|x-1|\Big]_2^{e+1}

    ...

    P.S. How do you make that line "|" longer?

    \color{red}|,\big|, \Big|, \bigg|, \Bigg|
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