Mathematics - Definite Integration Question with Solution | TestHub

Let $I={\int }_0^{1}xf(x)dx$. Using the property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$:$$I={\int }_0^1(1-x)f(1-x)dx$$ Adding the two expressions for $I$:$$2I={\int }_0^1[xf(x)+(1-x)f(1-x)]dx$$ From the given functional equation, $f(1-x)=\sin (\pi x)-f(x)$. Substitute this into the integral:$$2I={\int }_0^1[xf(x)+(1-x)(\sin (\pi x)-f(x))]dx$$ $$2I={\int }_0^1[xf(x)+(1-x)\sin (\pi x)-(1-x)f(x)]dx$$ $$2I={\int }_0^1[(2x-1)f(x)+(1-x)\sin (\pi x)]dx$$ Let $K={\int }_0^1(2x-1)f(x)dx$. Applying the property $x\to 1-x$ to $K$:$$K={\int }_0^1(2(1-x)-1)f(1-x)dx={\int }_0^1(1-2x)f(1-x)dx=-{\int }_0^1(2x-1)f(1-x)dx$$ Substitute $f(1-x)=\sin (\pi x)-f(x)$:$$K=-{\int }_0^1(2x-1)(\sin (\pi x)-f(x))dx=-{\int }_0^1(2x-1)\sin (\pi x)dx+{\int }_0^1(2x-1)f(x)dx$$ $$K=-{\int }_0^1(2x-1)\sin (\pi x)dx+K$$ This implies ${\int }_0^1(2x-1)\sin (\pi x)dx=0$. Let's verify this:$${\int }_0^1(2x-1)\sin (\pi x)dx={\int }_0^{1}2x\sin (\pi x)dx-{\int }_0^1\sin (\pi x)dx$$ For ${\int }_0^1\sin (\pi x)dx={\left[-\frac{\cos (\pi x)}{\pi }\right]}_0^1=-\frac{1}{\pi }(\cos \pi -\cos 0)=-\frac{1}{\pi }(-1-1)=\frac{2}{\pi }$. For ${\int }_0^{1}2x\sin (\pi x)dx$, use integration by parts with $u=2x,dv=\sin (\pi x)dx$:$${\left[-\frac{2x\cos (\pi x)}{\pi }\right]}_0^1-{\int }_0^1\left(-\frac{\cos (\pi x)}{\pi }\right)2dx=\left(-\frac{2(1)(-1)}{\pi }-0\right)+\frac{2}{\pi }{\left[\frac{\sin (\pi x)}{\pi }\right]}_0^1=\frac{2}{\pi }+0=\frac{2}{\pi }$$ So, ${\int }_0^1(2x-1)\sin (\pi x)dx=\frac{2}{\pi }-\frac{2}{\pi }=0$. This confirms $K=0$. Now substitute $K=0$ back into the expression for $2I$:$$2I=0+{\int }_0^1(1-x)\sin (\pi x)dx={\int }_0^1\sin (\pi x)dx-{\int }_0^{1}x\sin (\pi x)dx$$ We have ${\int }_0^1\sin (\pi x)dx=\frac{2}{\pi }$. For ${\int }_0^{1}x\sin (\pi x)dx$, use integration by parts with $u=x,dv=\sin (\pi x)dx$:$${\left[-\frac{x\cos (\pi x)}{\pi }\right]}_0^1-{\int }_0^1\left(-\frac{\cos (\pi x)}{\pi }\right)dx=\left(-\frac{1(-1)}{\pi }-0\right)+\frac{1}{\pi }{\left[\frac{\sin (\pi x)}{\pi }\right]}_0^1=\frac{1}{\pi }+0=\frac{1}{\pi }$$ So, $2I=\frac{2}{\pi }-\frac{1}{\pi }=\frac{1}{\pi }$. Thus, $I=\frac{1}{2\pi }$. The problem states $I=\frac{\pi }{k}$. Therefore, $\frac{1}{2\pi }=\frac{\pi }{k}⟹k=2{\pi }^2$.$k=2{\pi }^2\approx 2\times (3.14159{)}^2\approx 19.739$.