Q57 / 120 · VITEEE 2017 · Mathematics
$\int_{0}^{1}\left[\mathrm{f}(\mathrm{x}) \mathrm{g}^{\prime \prime}(\mathrm{x})-\mathrm{f}^{\prime \prime}(\mathrm{x}) \mathrm{g}(\mathrm{x})\right] \mathrm{d} \mathrm{x}$ is equal to :
$[$ Given $\mathrm{f}(0)=\mathrm{g}(0)=0]$
- A$\mathrm{f}(1) \mathrm{g}(1)-\mathrm{f}(1) \mathrm{g}^{\prime}(1)$
- B$\mathrm{f}(1) \mathrm{g}^{\prime}(1)+\mathrm{f}^{\prime}(1) \mathrm{g}(1)$
- Correct answerC$\mathrm{f}(1) \mathrm{g}^{\prime}(1)-\mathrm{f}^{\prime}(1) \mathrm{g}(1)$
- Dnone of these
Solution
Integrating by parts.
$$
\begin{array}{l}
\int f(x) g^{\prime \prime}(x) d x-\int f^{\prime \prime}(x) g(x) d x \\
=f(x) g^{\prime}(x)-\int f^{\prime}(x) g^{\prime}(x) d x \\
-f^{\prime}(x) g(x)+\int f^{\prime}(x) g^{\prime}(x) d x \\
=f(x) g^{\prime}(x)-f^{\prime}(x) g(x) \\
\text { Hence, } \int_{0}^{1} f(x) g^{\prime \prime}(x) d x-\int_{0}^{1} f^{\prime \prime}(x) g(x) d x \\
=f(1) g^{\prime}(1)-f^{\prime}(1) g(1)-f(0) g^{\prime}(0)+f^{\prime}(0) g(0) \\
=f(1) g^{\prime}(1)-f^{\prime}(1) g(1)
\end{array}
$$
Hence, the correct answer is option: 'C'.