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Thread: Definite Integral

  1. February 5th 2009, 04:12 PM #1
    Larrioto
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    Definite Integral

    h(1)=-2<br /> ,h'(1)=2<br /> ,h''(1)=3<br /> ,h(2)=6<br /> ,h'(2)=5<br /> ,h''(2)=13<br />

    Evaluate \int_1^2h''(u)du
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  2. February 5th 2009, 04:19 PM #2
    Danny
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    Quote Originally Posted by Larrioto View Post
    h(1)=-2<br /> ,h'(1)=2<br /> ,h''(1)=3<br /> ,h(2)=6<br /> ,h'(2)=5<br /> ,h''(2)=13<br />

    Evaluate \int_1^2h''(u)du
    \int_1^2h''(u)du = \left. h' (u)\,\right|_1^2 = h'(2) - h'(1)
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