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Let f be a twice differentiable function with f(7) = 2, f(8) = 7, f^{\prime} (7) = 2, and f^{\prime}(8) = 6. Evaluate the integral
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\[\int\limits_{7}^{8}xf \prime \prime(x)dx\]
Okay, so you'd do this by part, first let \[u=x\]\[\frac{ dv }{ dx }=f''(x)\]So your integral is \[I=\int\limits_{8}^{7}u \frac{ dv }{ dx }dx\]So now we find du/dx and integrate the dv/dx to get \[[uv]^8_7-\int\limits_{8}^{7}v \frac{ du }{ dx }dx\]So plugging in du/dx=1 and v=f'(x) since dv/dx=f''(x) we have\[[xf'(x)]^8_7-\int\limits_{8}^{7}f'(x)dx\] This goes down again to \[[xf'(x)]^8_7 - [f(x)]^8_7\]\[8f'(8)-7f'(7) - f(8)+f(7)\]Plug in the values for f'(8) etc and you get \[8.6 - 7.2 - 7+2= 48-14-7+2\] And your answer is 29. Hope this helped:)
It did thanks!
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