Question

Let f(x), g(x), h(x) be continous in [0, 2a] and satisfies $f(2a-x)=f\left(x\right),g(2a-x)=g\left(x\right),h\left(x\right)+h(2a-x)=3,f(2a-x)g(2a-x)=f\left(x\right)g\left(x\right)then{}^{}{\int }_0^{2a}f\left(x\right)g\left(x\right)h\left(x\right)dx=$

• A. ${\int }_0^{2a}f\left(x\right)g\left(x\right)dx$
• B. $3{\int }_0^{a}f\left(x\right)g\left(x\right)dx$
• C. $2{\int }_0^{a}f\left(x\right)g\left(x\right)dx$
• D. ${\int }_0^{a}f\left(x\right)g\left(x\right)dx$

Answer 1

The correct option is B $3{\int }_0^{a}f\left(x\right)g\left(x\right)dx$

$I={\int }_0^{2a}f\left(x\right)g\left(x\right)h\left(x\right)dx={\int }_0^{2a}f(2a-x)g(2a-x)h(2a-x)dx={\int }_0^{2a}f\left(x\right)g\left(x\right)[3-h(x\left)\right]dxI=\frac{3}{2}{\int }_0^{2a}f\left(x\right)g\left(x\right)dx=3{\int }_0^{a}f\left(x\right)g\left(x\right)dx$