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Math Help - i can't get past this simple problem

  1. #1
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    i can't get past this simple problem

    Show integral(a,b)(f(x+c)dx = integral(a+c,b+c)(f(x) dx.

    Let u = x+c, then du = dx so integral(a,b)(f(x+c)dx = integral(a+c,b+c)f(u)du. But when I substitute u back in, I get integral(a+c,b+c)(f(x+c)(du) which does not equal integral(a+c,b+c)(f(x)dx. What am I doing wrong?
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  2. #2
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    Quote Originally Posted by lord12 View Post
    Show integral(a,b)(f(x+c)dx = integral(a+c,b+c)(f(x) dx.

    Let u = x+c, then du = dx so integral(a,b)(f(x+c)dx = integral(a+c,b+c)f(u)du. But when I substitute u back in, I get integral(a+c,b+c)(f(x+c)(du) which does not equal integral(a+c,b+c)(f(x)dx. What am I doing wrong?
    When you substitute, u=x+c instead of f(x+c) you now have f(u).
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  3. #3
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    i still don't get it. Intuitively, both integrals should be equal. But if u = x+c and you are left with integral(a+c,b+c)(f(u))du, this equals integral(a+c,b+c)f(x+c)dx, not integral(a+c,b+c)(f(x)dx since u = x+c not u = x.
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  4. #4
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    Quote Originally Posted by lord12 View Post
    Show integral(a,b)(f(x+c)dx = integral(a+c,b+c)(f(x) dx.

    Let u = x+c, then du = dx so integral(a,b)(f(x+c)dx = integral(a+c,b+c)f(u)du. But when I substitute u back in, I get integral(a+c,b+c)(f(x+c)(du) which does not equal integral(a+c,b+c)(f(x)dx. What am I doing wrong?
    As TPH said:
    \int_a^bf(x + c)~dx

    Let u = x + c. Then the lower bound on u is u_- = a + c and the upper bound on u is u_+ = b + c. Thus

    \int_a^bf(x + c)~dx = \int_{a + c}^{b + c}f(u)~du
    as you desire.

    Once you change the variable in the integrand, you need to do the same variable change in the limits of the integral to be consistent.

    Now you can change the dummy variable u back to x:
    \int_a^bf(x + c)~dx = \int_{a + c}^{b + c}f(u)~du = \int_{a + c}^{b + c}f(x)~dx
    but note again that the integrand has been rewritten so that the arguments of f are different.

    -Dan
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  5. #5
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    thanks a lot guys! I was just confused a bit.
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