Substitution Rule for Definite Integrals — Question 1

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Question 1

Evaluate the definite integral ∫01x1+x2dx.\int_{0}^{1} \frac{x}{\sqrt{1+x^2}}\,dx.

Original worksheet page 1: question and worked solution for 5-8-001
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Question 1 - Solution

Use substitution to simplify the integrand.

Let u=1+x2.u=1+x^2. Then du=2xdx⇒xdx=12du.du=2x\,dx \quad\Rightarrow\quad x\,dx=\frac12\,du.

Change the limits of integration. When x=0x=0, u=1u=1. When x=1x=1, u=2u=2.

Substitute into the integral: ∫01x1+x2dx=12∫12u−1/2du.\int_{0}^{1} \frac{x}{\sqrt{1+x^2}}\,dx = \frac12\int_{1}^{2} u^{-1/2}\,du.

Integrate: 12∫u−1/2du=12⋅2u1/2=u.\frac12\int u^{-1/2}\,du = \frac12\cdot 2u^{1/2} = \sqrt{u}.

Apply the new limits: u|12=2−1.\sqrt{u}\Big|_{1}^{2} = \sqrt{2}-1.

2−1\boxed{\sqrt{2}-1}

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