Q88 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
If $f(x)$ and $g(x)$ are integrable functions then $\left[\int f(x)dx\right]\left[\int g(x)dx\right] = $
- A$\int \left[f(x)g'(x) + f'(x)g(x)\right]dx$
- B$\int \left[f(x)g'(x) - f'(x)g(x)\right]dx$
- Correct answerC$\int \left[f(x)\int g(x)dx + g(x)\int f(x)dx\right]dx$
- D$\int \left[f(x)\int g(x)dx - g(x)\int f(x)dx\right]dx$
Solution
Let $u = \int f(x)dx$ and $v = \int g(x)dx$.
Differentiating with respect to $x$, we get $\dfrac{du}{dx} = f(x)$ and $\dfrac{dv}{dx} = g(x)$.
Using the product rule of differentiation, $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$.
Substituting the expressions for $u$, $v$, $\dfrac{du}{dx}$, and $\dfrac{dv}{dx}$, we get $\dfrac{d}{dx}\left[\left(\int f(x)dx\right)\left(\int g(x)dx\right)\right] = \left(\int f(x)dx\right)g(x) + \left(\int g(x)dx\right)f(x)$.
Integrating both sides with respect to $x$, we obtain $\left[\int f(x)dx\right]\left[\int g(x)dx\right] = \int \left[f(x)\int g(x)dx + g(x)\int f(x)dx\right]dx$.
Answer: $\int \left[f(x)\int g(x)dx + g(x)\int f(x)dx\right]dx$
Differentiating with respect to $x$, we get $\dfrac{du}{dx} = f(x)$ and $\dfrac{dv}{dx} = g(x)$.
Using the product rule of differentiation, $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$.
Substituting the expressions for $u$, $v$, $\dfrac{du}{dx}$, and $\dfrac{dv}{dx}$, we get $\dfrac{d}{dx}\left[\left(\int f(x)dx\right)\left(\int g(x)dx\right)\right] = \left(\int f(x)dx\right)g(x) + \left(\int g(x)dx\right)f(x)$.
Integrating both sides with respect to $x$, we obtain $\left[\int f(x)dx\right]\left[\int g(x)dx\right] = \int \left[f(x)\int g(x)dx + g(x)\int f(x)dx\right]dx$.
Answer: $\int \left[f(x)\int g(x)dx + g(x)\int f(x)dx\right]dx$
Hence, the correct answer is option: 'C'.