Question

If $f"\left(x\right)=kin[0,a],then{\int }_0^{a}f\left(x\right)dx-{\left\{\begin{array}{c}xf\left(x\right)-\frac{x^2}{2!}{f}^{\prime }\left(x\right)+\frac{x^3}{3!}f"\left(x\right)\end{array}\right\}}_0^a$ is

• A. $\frac{-k{a}^4}{12}$
• B. $\frac{k{a}^4}{24}$
• C. $\frac{-k{a}^4}{24}$
• D. $\frac{k{a}^4}{12}$

Answer 1

The correct option is C $\frac{-k{a}^4}{24}$

Let $I={\int }_0^{a}f\left(x\right)dx-{\left\{xf\right(x)dx-\frac{x^2}{2!}{f}^{\prime }(x)+\frac{x^3}{3!}{f}^{\prime \prime }(x\left)\right\}}_0^a.....\left(1\right)Let{I}_1={\int }_0^{a}f\left(x\right)dx.={\left[xf\right(x\left)\right]}_0^a-{\int }_0^{a}x{f}^{\prime }\left(x\right)dx={\left[xf\right(x\left)\right]}_0^a-{\left[\frac{x^2}{2!}{f}^{\prime }\right(x\left)\right]}_0^a+{\int }_0^a\frac{x^2}{2!}{f}^{\prime \prime }\left(x\right)dx={\left[xf\right(x)-\frac{x^2}{2!}{f}^{\prime }(x\left)\right]}_0^a+{\left[\frac{x^3}{3!}{f}^{\prime \prime }\right(x\left)\right]}_0^a-{\int }_0^a\frac{x^3}{3!}{f}^{\prime \prime }\left(x\right)dx={\left[xf\right(x)-\frac{x^2}{2!}{f}^{\prime }(x)+\frac{x^3}{3!}{f}^{\prime \prime }(x\left)\right]}_0^a-{\left[\frac{x^4}{4!}{f}^{\prime \prime \prime }\right(x\left)\right]}_0^a+{\int }_0^a\frac{x^4}{4!}{f}^{\prime \prime \prime \prime }\left(x\right)dx={[xf-\frac{x^2}{2!}{f}^{\prime }(x)+\frac{x^3}{3!}{f}^{\prime \prime }(x\left)\right]}_0^a-{\left[\frac{x^4}{4!}{f}^{\prime \prime \prime }\right(x\left)\right]}_0^a.....\left(2\right)$
Substituting (2) is (1) use get
$I=\frac{-K{a}^4}{24}$