CBSE 2025 · Region 2 · Set 2 · Q31 · 3 marks
f and g are continuous functions on interval $\displaystyle [\mathrm{a}, \mathrm{b}]$. Given that $\displaystyle \mathrm{f}(\mathrm{a}-x)=\mathrm{f}(x)$ and $\displaystyle \mathrm{g}(x)+\mathrm{g}(\mathrm{a}-x)=\mathrm{a}$, show that $\displaystyle \int_{0}^{\mathrm{a}} \mathrm{f}(x) \mathrm{g}(x) \mathrm{d} x=\frac{\mathrm{a}}{2} \int_{0}^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x$.
Marking-scheme solution
\[\begin{aligned}
& I=\int_{0}^{\mathrm{a}} \mathrm{f}(x) \mathrm{g}(x) \mathrm{d} x \\
& =\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-x) \mathrm{g}(\mathrm{a}-x) \mathrm{d} x \\
& =\int_{0}^{\mathrm{a}} \mathrm{f}(x)[\mathrm{a}-\mathrm{g}(x)] \mathrm{d} x \\
& I=\mathrm{a} \int_{0}^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x-\int_{0}^{\mathrm{a}} \mathrm{f}(x) \mathrm{g}(x) \mathrm{d} x \\
& \text { Adding, we get } I=\frac{\mathrm{a}}{2} \int_{0}^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x
\end{aligned}
\]
IntegralsSome Properties of Definite IntegralsApplyshort_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.