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Math Help - Fundamental Theorem of Calculus

  1. #1
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    Fundamental Theorem of Calculus

    Need some help with FTC problems. My calculus textbook defines the FTC in two parts:

    Definition Part 1:
    If f is a continuous function on [a,b], then the function g defined by g(x) = \int_a^x f(t)dt where a\leq x\leq b
    is continous on [a.b] and differentiable on (a,b), and g'(x) = f(x)

    Definition Part 2:
    If f is a continuous function on [a,b], then \int_a^b f(x)dx=F(b)-F(a) where F is any antiderivative of f, that is, a function such that F'=f

    Whew, here are the problems I'm confused on...

    Problem 1. Using Part 1 definition, find the derivative of the function:
    y=\int_{1-3x}^1 \frac{u^3}{1+u^2}du

    Problem 2. Using Part 2 defintion, evaluate the integral:
    \int_0^2 x(2+x^5)dx

    Thanks a lot.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by c_323_h
    Need some help with FTC problems. My calculus textbook defines the FTC in two parts:

    Definition Part 1:
    If f is a continuous function on [a,b], then the function g defined by g(x) = \int_a^x f(t)dt where a\leq x\leq b
    is continous on [a.b] and differentiable on (a,b), and g'(x) = f(x)

    Definition Part 2:
    If f is a continuous function on [a,b], then \int_a^b f(x)dx=F(b)-F(a) where F is any antiderivative of f, that is, a function such that F'=f

    Whew, here are the problems I'm confused on...

    Problem 1. Using Part 1 definition, find the derivative of the function:
    y=\int_{1-3x}^1 \frac{u^3}{1+u^2}du

    Problem 2. Using Part 2 defintion, evaluate the integral:
    \int_0^2 x(2+x^5)dx

    Thanks a lot.
    Problem 1
    g(x)=y
    f(x)= \frac{u^3}{1+u^2}
    Problem 2
    Intregrate the integrand and put the limits

    Keep Smiling
    Malay
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  3. #3
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    Quote Originally Posted by malaygoel
    Problem 1
    g(x)=y
    f(x)= \frac{u^3}{1+u^2}
    Problem 2
    Intregrate the integrand and put the limits

    Keep Smiling
    Malay
    hmmm....i don't know how to integrate the integrand. that is why i am asking
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  4. #4
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    Quote Originally Posted by c_323_h

    Problem 1. Using Part 1 definition, find the derivative of the function:
    y=\int_{1-3x}^1 \frac{u^3}{1+u^2}du
    That is same as,
    y=-\int_1^{1-3x}\frac{u^3}{1+u^2}du
    Consider a function,
    F(x)=\int_1^x\frac{u^3}{1+u^2}du
    by, FTC we have,
    F'(x)=\frac{x^3}{1+x^2}
    Now note that,
    y=-(F\circ 1-3x)(x)=-F(1-3x)
    A function composition!
    Then by the Chain Rule,
    <br />
y'=-F'(1-3x)(-3)=3F'(1-3x)
    But,
    F'=\frac{x^3}{1+x^2}
    Thus,
    y'=\frac{3(1-3x)^3}{1+(1-3x)^2}
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker
    That is same as,
    y=-\int_1^{1-3x}\frac{u^3}{1+u^2}du
    Consider a function,
    F(x)=\int_1^x\frac{u^3}{1+u^2}du
    by, FTC we have,
    F'(x)=\frac{x^3}{1+x^2}
    Now note that,
    y=-(F\circ 1-3x)(x)=-F(1-3x)
    A function composition!
    Then by the Chain Rule,
    <br />
y'=-F'(1-3x)(-3)=3F'(1-3x)
    But,
    F'=\frac{x^3}{1+x^2}
    Thus,
    y'=\frac{3(1-3x)^3}{1+(1-3x)^2}
    Can I just butt in and say that I've never seen anything like that and Dude! That was TOTALLY awesome!

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark
    Can I just butt in and say that I've never seen anything like that and Dude! That was TOTALLY awesome!

    -Dan
    I know, i am the best.
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