∫ 1 4 x f ″ ( x ) d x u = x d v = f ″ ( x ) ⋅ ( d x ) d u = d x v = f ′ ( x ) ∫ 1 4 x f ″ ( x ) d x = x ⋅ f ′ ( x ) | 1 4 − ∫ 1 4 f ′ ( x ) d x = x ⋅ f ′ ( x ) − f ( x ) | 1 4 = ( 4 ⋅ f ′ ( 4 ) − f ( 4 ) ) − ( 1 ⋅ f ′ ( 1 ) − f ( 1 ) ) = ( 4 ⋅ 3 − 7 ) − ( 5 − 2 ) = ( 12 − 7 ) − ( 3 ) = ( 5 ) − ( 3 ) = 2 {\displaystyle {\begin{aligned}\int _{1}^{4}xf''(x)dx\\[2ex]u=x\qquad dv=f''(x)\cdot \;(dx)\qquad \\[2ex]du=dx\qquad v=f'(x)\qquad \\[2ex]\int _{1}^{4}xf''(x)dx&=x\cdot \;f'(x){\bigg |}_{1}^{4}-\int _{1}^{4}f'(x)dx\\[2ex]&=x\cdot \;f'(x)-f(x){\bigg |}_{1}^{4}\\[2ex]&=(4\cdot \;f'(4)-f(4))-(1\cdot \;f'(1)-f(1))\\[2ex]&=(4\cdot \;3-7)-(5-2)\\[2ex]&=(12-7)-(3)\\[2ex]&=(5)-(3)\\[2ex]&=2\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{1}^{4}xf''(x)dx\\[2ex]u=x\qquad dv=f''(x)\cdot \;(dx)\qquad \\[2ex]du=dx\qquad v=f'(x)\qquad \\[2ex]\int _{1}^{4}xf''(x)dx&=x\cdot \;f'(x){\bigg |}_{1}^{4}-\int _{1}^{4}f'(x)dx\\[2ex]&=x\cdot \;f'(x)-f(x){\bigg |}_{1}^{4}\\[2ex]&=(4\cdot \;f'(4)-f(4))-(1\cdot \;f'(1)-f(1))\\[2ex]&=(4\cdot \;3-7)-(5-2)\\[2ex]&=(12-7)-(3)\\[2ex]&=(5)-(3)\\[2ex]&=2\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/11b86cf91e79bf1047b86ef77d5116e7018c16a5)