CUET UG 2025 — Mathematics Calculus

Given that $f(a-x) = f(x)$f(a−x)=f(x), this indicates the function is symmetric about $x = \frac{a}{2}$x=2a​.

Let $\displaystyle I = \int_0^a xf(x)dx$I=∫0a​xf(x)dx

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Using the property $\displaystyle \int_0^a g(x)dx = \int_0^a g(a-x)dx$∫0a​g(x)dx=∫0a​g(a−x)dx, replace $x$x with $a-x$a−x:

$$I = \int_0^a (a-x)f(a-x)dx$$I=∫0a​(a−x)f(a−x)dx

Since $f(a-x) = f(x)$f(a−x)=f(x):

$$I = \int_0^a (a-x)f(x)dx$$I=∫0a​(a−x)f(x)dx

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Expanding the integral:

$$I = \int_0^a af(x)dx - \int_0^a xf(x)dx$$I=∫0a​af(x)dx−∫0a​xf(x)dx

$$I = a\int_0^a f(x)dx - I$$I=a∫0a​f(x)dx−I

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Solving for $I$I:

$$I + I = a\int_0^a f(x)dx$$I+I=a∫0a​f(x)dx

$$2I = a\int_0^a f(x)dx$$2I=a∫0a​f(x)dx

$$I = \frac{a}{2}\int_0^a f(x)dx$$I=2a​∫0a​f(x)dx

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Therefore, $\displaystyle \int_0^a xf(x)dx = \frac{a}{2}\int_0^a f(x)dx$∫0a​xf(x)dx=2a​∫0a​f(x)dx