1 markMathematicsIntegration and its applicationsIntegration techniquesModerateabout 2 min

IIT JAM 2021 Question 44: Integration and its applications

The value of the integral $\displaystyle\int_0^9 \dfrac{x - 1}{1 + \sqrt{x}}\,dx$ is _______ (in integer).

Answer and worked solution

Answer: 9

Let $u = \sqrt{x}$ , so $x = u^2$ and $dx = 2u\,du$ , with $x: 0 \to 9$ giving $u: 0 \to 3$ . Then $\dfrac{x-1}{1+\sqrt{x}} = \dfrac{(u-1)(u+1)}{1+u} = u - 1$ .

The integral becomes $\int_0^3 (u-1)\,2u\,du = 2\int_0^3 (u^2 - u)\,du = 2\left[9 - \tfrac92\right] = 9$ . The answer is $9$ .

Plain-text version

The value of the integral ∫₀⁹ (x - 1)/(1 + √x) dx is _ (in integer).

Also written as: The value of the integral integral_0^9 (x - 1)/(1 + sqrt(x)) dx is _ (in integer).

Integration and its applications has come up 43 times in 6 exams since 2007.

What this question tests

This question is on Integration techniques, part of Integration and its applications in Mathematics.

Substitution, integration by parts, partial fractions and improper integrals.

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