38. if f(0)=g(0)=0 and f and g are continuous, show that (int_{0}^{a}f(x)g(x)dx = f(a)g(a)-f(a)g(a)+int_{0}^{…

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula for $\int_{0}^{a}u\mathrm{d}v=uv|{0}^{a}-\int{0}^{a}v\mathrm{d}u$. Let $u = f(x)$ and $\mathrm{d}v=g''(x)\mathrm{d}x$. Then $\mathrm{d}u = f'(x)\mathrm{d}x$ and $v = g'(x)$. So, $\int_{0}^{a}f(x)g''(x)\mathrm{d}x=f(x)g'(x)|{0}^{a}-\int{0}^{a}f'(x)g'(x)\mathrm{d}x$. Evaluating $f(x)g'(x)|_{0}^{a}$, we get $f(a)g'(a)-f(0)g'(0)$. Since $f(0) = 0$, it simplifies to $f(a)g'(a)$.

Step2: Apply integration - by - parts again

Now, for $\int_{0}^{a}f'(x)g'(x)\mathrm{d}x$, let $u = f'(x)$ and $\mathrm{d}v=g'(x)\mathrm{d}x$. Then $\mathrm{d}u = f''(x)\mathrm{d}x$ and $v = g(x)$. So, $\int_{0}^{a}f'(x)g'(x)\mathrm{d}x=f'(x)g(x)|{0}^{a}-\int{0}^{a}f''(x)g(x)\mathrm{d}x$. Evaluating $f'(x)g(x)|_{0}^{a}$, we get $f'(a)g(a)-f'(0)g(0)$. Since $g(0) = 0$, it simplifies to $f'(a)g(a)$.

Step3: Combine the results

Substituting the results of the two - step integration - by - parts into the original equation: $\int_{0}^{a}f(x)g''(x)\mathrm{d}x=f(a)g'(a)-\left(f'(a)g(a)-\int_{0}^{a}f''(x)g(x)\mathrm{d}x\right)=f(a)g'(a)-f'(a)g(a)+\int_{0}^{a}f''(x)g(x)\mathrm{d}x$.

Answer:

The given equation $\int_{0}^{a}f(x)g''(x)\mathrm{d}x=f(a)g'(a)-f'(a)g(a)+\int_{0}^{a}f''(x)g(x)\mathrm{d}x$ is shown through two - step application of the integration - by - parts formula.