We use integration by parts and set u = f ′ ( x ) {\displaystyle u=f'(x)} and d v = f ″ ( x ) d x {\displaystyle dv=f''(x)dx} . Then
∫ 1 2 f ′ ( x ) f ″ ( x ) d x = [ f ′ ( x ) 2 ] 1 2 − ∫ 1 2 f ″ ( x ) f ′ ( x ) d x {\displaystyle \int _{1}^{2}f'(x)f''(x)dx=\left[f'(x)^{2}\right]_{1}^{2}-\int _{1}^{2}f''(x)f'(x)dx}
Note that the integral on the left and right are the same and thus we can rewrite this as
2 ∫ 1 2 f ′ ( x ) f ″ ( x ) d x = [ f ′ ( x ) 2 ] 1 2 = 3 2 − 2 2 = 9 − 4 = 5 {\displaystyle 2\int _{1}^{2}f'(x)f''(x)dx=\left[f'(x)^{2}\right]_{1}^{2}=3^{2}-2^{2}=9-4=5}
Dividing by 2 leads to ∫ 1 2 f ′ ( x ) f ″ ( x ) d x = 5 2 {\displaystyle \int _{1}^{2}f'(x)f''(x)dx={\frac {5}{2}}}
and 
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Then
![{\displaystyle \int _{1}^{2}f'(x)f''(x)dx=\left[f'(x)^{2}\right]_{1}^{2}-\int _{1}^{2}f''(x)f'(x)dx}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/39534036e09dd2e44d6da0b1763c672614657a3a)
![{\displaystyle 2\int _{1}^{2}f'(x)f''(x)dx=\left[f'(x)^{2}\right]_{1}^{2}=3^{2}-2^{2}=9-4=5}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7a2c7aeefc9296b294ce300a779f50fc0d943872)
