If f(x) and g(x) are integrable functions then ≤ft[ f(x) d x]≤ft[ g(x) d x]=

Correct Answer is : ∫[f(x)∫g(x)dx+g(x)∫f(x)dx]dx≤ft[f(x) g(x) d x+g(x) f(x) d x] d x∫[f(x)∫g(x)dx+g(x)∫f(x)dx]dx

≤ft[f(x) g^{}(x)+f^{}(x) g(x)] d x

≤ft[f(x) g^{}(x)-f^{}(x) g(x)] d x

≤ft[f(x) g(x) d x+g(x) f(x) d x] d x

≤ft[f(x) g(x) d x-g(x) f(x) d x] d x

Correct Answer is : ∫[f(x)∫g(x)dx+g(x)∫f(x)dx]dx≤ft[f(x) g(x) d x+g(x) f(x) d x] d x∫[f(x)∫g(x)dx+g(x)∫f(x)dx]dx

Test Index

MHTCET 15 Apr 2026 Shift 1 Paper

Question : 129 of 150

Marks: +1, -0

f(x)

g(x)

\left[\int f(x) d x\right]\left[\int g(x) d x\right]=

Solution:

Concept:

Integration as antiderivative and the product rule of differentiation link the product of two integrals.

Explanation:

Let

F(x) = \int f(x)\,dx

and

G(x) = \int g(x)\,dx

Then

F'(x) = f(x)

and

G'(x) = g(x)

The given expression is

F(x)G(x)

Using the product rule,

\frac{d}{dx}[F(x)G(x)] = F'(x)G(x) + F(x)G'(x)

Substituting

F'(x) = f(x)

and

G'(x) = g(x)

, we get:

\frac{d}{dx}[F(x)G(x)] = f(x)\int g(x)\,dx + g(x)\int f(x)\,dx

Integrating both sides with respect to

x

gives:

F(x)G(x) = \int\left[f(x)\int g(x)\,dx + g(x)\int f(x)\,dx\right]dx

This matches option C.

Answer:

\displaystyle \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

Therefore, the correct option is Option C.

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