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Math Help - The First Fundamental Theorem,definite integral.......

  1. #1
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    The First Fundamental Theorem,definite integral.......

    Hello,
    please try to solve these questions.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by m777 View Post
    Hello,
    please try to solve these questions.
    Question #1

    Show that the area under y=x^3 over the interval [0,b] is \frac{b^4}{4}


    This is asking for the definite integral:

    I=\int_0^b \,x^3\, dx

    You should know that x^3 is the derivative of x^4/4, and so:

    I=\int_0^b \,x^3\, dx=\left[x^4/4\right]_0^b=b^4/4-0^4/4=b^4/4

    RonL
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  3. #3
    Grand Panjandrum
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    Question #2

    (a) Evaluate the definite integral using “The First Fundamental Theorem” of calculus.

    \int_1^2 \left( \frac{1}{x^3}-\frac{2}{x^2}+\frac{1}{x^4}\right)\,dx

    This is asking you to spot what the integrand is the derivative of then by the
    first fundamental theorem of calc the integral is the difference of the
    derivative at the end points.

    Now:

     \frac{1}{x^3}-\frac{2}{x^2}+\frac{1}{x^4} = \frac{d}{dx}\left(-\frac{1}{2x^2}+\frac{2}{x}-\frac{1}{3x^3}\right)

    So:

    \int_1^2 \left( \frac{1}{x^3}-\frac{2}{x^2}+\frac{1}{x^4}\right)\,dx = \left[-\frac{1}{2x^2}+\frac{2}{x}-\frac{1}{3x^3}\right]_1^2 =-\frac{1}{8}+1-\frac{1}{24}+\frac{1}{2}-2+\frac{1}{3} = -\frac{1}{3}

    RonL
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