Math  /  Calculus

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Use a change of variables to evaluate the definite integral 01(2x1x2)dx\int_{0}^{1}\left(2 x \sqrt{1-x^{2}}\right) d x. Leave your final answer as a fraction. 01(2x1x2)dx=\int_{0}^{1}\left(2 x \sqrt{1-x^{2}}\right) d x= \square

Studdy Solution

STEP 1

1. We are given the definite integral 01(2x1x2)dx\int_{0}^{1}\left(2 x \sqrt{1-x^{2}}\right) d x.
2. We need to use a change of variables to evaluate this integral.
3. The final answer should be left as a fraction.

STEP 2

1. Choose an appropriate substitution for the variable xx.
2. Determine the new limits of integration.
3. Substitute and transform the integral.
4. Evaluate the transformed integral.
5. Simplify the result to a fraction.

STEP 3

Choose a substitution. Let x=sinθ x = \sin \theta . Then, dx=cosθdθ dx = \cos \theta \, d\theta .

STEP 4

Determine the new limits of integration: - When x=0 x = 0 , sinθ=0 \sin \theta = 0 implies θ=0 \theta = 0 . - When x=1 x = 1 , sinθ=1 \sin \theta = 1 implies θ=π2 \theta = \frac{\pi}{2} .

STEP 5

Substitute into the integral: 012x1x2dx=0π22sinθ1sin2θcosθdθ \int_{0}^{1} 2x \sqrt{1-x^2} \, dx = \int_{0}^{\frac{\pi}{2}} 2\sin \theta \sqrt{1-\sin^2 \theta} \cdot \cos \theta \, d\theta
Simplify using the identity 1sin2θ=cosθ\sqrt{1-\sin^2 \theta} = \cos \theta: =0π22sinθcos2θdθ = \int_{0}^{\frac{\pi}{2}} 2\sin \theta \cos^2 \theta \, d\theta

STEP 6

Use the identity cos2θ=1+cos2θ2\cos^2 \theta = \frac{1+\cos 2\theta}{2} to simplify: =0π22sinθ1+cos2θ2dθ = \int_{0}^{\frac{\pi}{2}} 2\sin \theta \cdot \frac{1+\cos 2\theta}{2} \, d\theta =0π2sinθ(1+cos2θ)dθ = \int_{0}^{\frac{\pi}{2}} \sin \theta (1+\cos 2\theta) \, d\theta
Separate the integral: =0π2sinθdθ+0π2sinθcos2θdθ = \int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta + \int_{0}^{\frac{\pi}{2}} \sin \theta \cos 2\theta \, d\theta
Evaluate each integral separately:
1. 0π2sinθdθ=[cosθ]0π2=1\int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta = [-\cos \theta]_{0}^{\frac{\pi}{2}} = 1

2. Use integration by parts for 0π2sinθcos2θdθ\int_{0}^{\frac{\pi}{2}} \sin \theta \cos 2\theta \, d\theta: - Let u=sinθ u = \sin \theta , dv=cos2θdθ dv = \cos 2\theta \, d\theta - Then du=cosθdθ du = \cos \theta \, d\theta , v=12sin2θ v = \frac{1}{2}\sin 2\theta
=[12sinθsin2θ]0π2120π2sin2θcosθdθ = \left[\frac{1}{2} \sin \theta \sin 2\theta \right]_{0}^{\frac{\pi}{2}} - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin 2\theta \cos \theta \, d\theta
The second integral evaluates to zero due to symmetry.

STEP 7

Combine the results: =1+0=1 = 1 + 0 = 1
The value of the definite integral is:
1 \boxed{1}

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