Question

If f and g are continuous on [0, a] and satisfy f(x) = f(a - x) and g(x) + g(x - a) = 2, then ${\int }_0^{a}f\left(x\right)g\left(x\right)\mathrm{dx}$is equal to

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

Answer 1

Answer: (A), (B), (C)

The correct options are
A ${\int }_0^{a}f\left(x\right)\mathrm{dx}$
B ${\int }_0^{a}f(a-x)\mathrm{dx}$
C ${\int }_0^{a}f(a-x)g(a-x)\mathrm{dx}$

$\because {\int }_0^{a}f\left(x\right)\mathrm{dx}={\int }_0^{a}f(a-x)\mathrm{dx}\therefore {\int }_0^{a}f\left(x\right)g\left(x\right)\mathrm{dx}={\int }_0^{a}f(a-x)g(a-x)\mathrm{dx}\Rightarrow {\int }_0^{a}f\left(x\right)g\left(x\right)\mathrm{dx}={\int }_0^{a}f(a-x)\left(2-g\left(x\right)\right\}\mathrm{dx}\Rightarrow {\int }_0^{a}f\left(x\right)g\left(x\right)\mathrm{dx}=2{\int }_0^{a}f\left(x\right)\mathrm{dx}-{\int }_0^{a}f\left(x\right)g\left(x\right)\mathrm{dx}\Rightarrow 2{\int }_0^{a}f\left(x\right)g\left(x\right)\mathrm{dx}=2{\int }_0^{a}f\left(x\right)\mathrm{dx}\Rightarrow {\int }_0^{a}f\left(x\right)g\left(x\right)\mathrm{dx}={\int }_0^{a}f\left(x\right)\mathrm{dx}$