Mathematics - Definite Integration Question with Solution | TestHub

From property (i), $f(x)+f(2-x)=2$. Integrating from 0 to 2:${\int }_0^2(f(x)+f(2-x))dx={\int }_0^{2}2dx=4$ Using King's rule for the second term, ${\int }_0^{2}f(2-x)dx={\int }_0^{2}f(x)dx$. Thus, $2{\int }_0^{2}f(x)dx=4⟹{\int }_0^{2}f(x)dx=2$. Next, consider ${\int }_0^{4}f(x)dx={\int }_0^{2}f(x)dx+{\int }_2^{4}f(x)dx$. For the second integral, let $x=2+t$. ${\int }_2^{4}f(x)dx={\int }_0^{2}f(2+t)dt$. From property (i), $f(x)+f(2-x)=2$. Replacing $x$ with $2+t$, we get $f(2+t)+f(2-(2+t))=2⟹f(2+t)+f(-t)=2$. So, ${\int }_0^{2}f(2+t)dt={\int }_0^2(2-f(-t))dt=4-{\int }_0^{2}f(-t)dt$. Using property (ii), $f(-t)=f(-t+4)=f(4-t)$. So ${\int }_0^{2}f(-t)dt={\int }_0^{2}f(4-t)dt$. Let $u=4-t$, then $du=-dt$. When $t=0,u=4$; when $t=2,u=2$. ${\int }_0^{2}f(4-t)dt={\int }_4^{2}f(u)(-du)={\int }_2^{4}f(u)du$. Substituting back, ${\int }_2^{4}f(x)dx=4-{\int }_2^{4}f(x)dx⟹2{\int }_2^{4}f(x)dx=4⟹{\int }_2^{4}f(x)dx=2$. Therefore, ${\int }_0^{4}f(x)dx=2+2=4$. Finally, $I={\int }_0^{10}f(x)dx$. Since $f(x)$ has period 4, $I={\int }_0^{2\times 4+2}f(x)dx=2{\int }_0^{4}f(x)dx+{\int }_8^{10}f(x)dx$. Using periodicity, ${\int }_8^{10}f(x)dx={\int }_0^{2}f(x)dx$. So, $I=2(4)+2=8+2=10$.