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

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

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

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

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

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

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

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

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

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

    Thanks a lot.
    Problem 1
    g(x)=y
    f(x)=$\displaystyle \frac{u^3}{1+u^2}$
    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)=$\displaystyle \frac{u^3}{1+u^2}$
    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:
    $\displaystyle y=\int_{1-3x}^1 \frac{u^3}{1+u^2}du$
    That is same as,
    $\displaystyle y=-\int_1^{1-3x}\frac{u^3}{1+u^2}du$
    Consider a function,
    $\displaystyle F(x)=\int_1^x\frac{u^3}{1+u^2}du$
    by, FTC we have,
    $\displaystyle F'(x)=\frac{x^3}{1+x^2}$
    Now note that,
    $\displaystyle y=-(F\circ 1-3x)(x)=-F(1-3x)$
    A function composition!
    Then by the Chain Rule,
    $\displaystyle
    y'=-F'(1-3x)(-3)=3F'(1-3x)$
    But,
    $\displaystyle F'=\frac{x^3}{1+x^2}$
    Thus,
    $\displaystyle y'=\frac{3(1-3x)^3}{1+(1-3x)^2}$
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker
    That is same as,
    $\displaystyle y=-\int_1^{1-3x}\frac{u^3}{1+u^2}du$
    Consider a function,
    $\displaystyle F(x)=\int_1^x\frac{u^3}{1+u^2}du$
    by, FTC we have,
    $\displaystyle F'(x)=\frac{x^3}{1+x^2}$
    Now note that,
    $\displaystyle y=-(F\circ 1-3x)(x)=-F(1-3x)$
    A function composition!
    Then by the Chain Rule,
    $\displaystyle
    y'=-F'(1-3x)(-3)=3F'(1-3x)$
    But,
    $\displaystyle F'=\frac{x^3}{1+x^2}$
    Thus,
    $\displaystyle y'=\frac{3(1-3x)^3}{1+(1-3x)^2}$
    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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