Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals. $$\int_{0}^{1} x \sqrt{1-x^{2}} d x$$

Short answer

Question: Evaluate the definite integral $$\int_{0}^{1} x \sqrt{1-x^{2}} d x$$. Answer: The definite integral evaluates to $$\frac{\pi}{4}$$.

Step-by-step solution

Choose the substitution

We will make the substitution x = sin(θ). To find the differential dx, differentiate both sides with respect to θ: $$\frac{d x}{d \theta} = \cos (\theta)$$ And then multiply by dθ: $$d x = \cos (\theta) d\theta$$ And lastly, change the limits of integration accordingly: x = 0: θ = arcsin(0) = 0 x = 1: θ = arcsin(1) = \frac{\pi}{2}

Replace dx and change the limits of integration

Substitute x and dx into the original definite integral: $$\int_{0}^{1} x \sqrt{1-x^{2}} d x = \int_{0}^{\frac{\pi}{2}} (\sin \theta) \sqrt{1-\sin^2 \theta} \cos(\theta) d \theta$$

Simplify the integral

The square root term can be simplified using the identity 1 - sin^2(θ) = cos^2(θ): $$\int_{0}^{\frac{\pi}{2}} (\sin\theta) \sqrt{\cos^2\theta} \cos(\theta) d\theta$$ Now, we know that the square root of cos^2(θ) is |cos(θ)|. Since θ is between 0 and π/2, cos(θ) is positive and we can drop the absolute value: $$\int_{0}^{\frac{\pi}{2}} (\sin\theta) \cos(\theta) \cos(\theta) d\theta$$ Combine the cos(θ) terms: $$\int_{0}^{\frac{\pi}{2}} (\sin\theta) \cos^2(\theta) d\theta$$

Evaluate the integral

To evaluate the integral $$\int_{0}^{\frac{\pi}{2}} (\sin\theta) \cos^2(\theta) d\theta$$, we can use integration by parts. Choose u = sin(θ) (so du = cos(θ)dθ) and dv = cos^2(θ)dθ (so v = integral of cos^2(θ)dθ which is (1/2)(θ + sin(θ)cos(θ))): Using the equation $$\int u dv = uv - \int v du$$, $$\int_{0}^{\frac{\pi}{2}} (\sin\theta) \cos^2(\theta) d\theta = (\sin\theta)\left(\frac{1}{2}\right)(\theta + \sin(\theta)\cos(\theta)) - \int_{0}^{\frac{\pi}{2}} (\frac{1}{2})(\theta + \sin(\theta)\cos(\theta)) \cos(\theta) d\theta$$ Now, split the integral into two parts and evaluate: $$= \left[ \frac{1}{2}\theta\sin\theta + \frac{1}{4}\sin^2\theta\cos\theta \right]_{0}^{\frac{\pi}{2}}$$

Revert the substitution and solve to find the result

When evaluating the integral, we'll keep in mind that sin(0) = 0, sin(π/2) = 1, cos(0) = 1, and cos(π/2) = 0: $$= \left[ \frac{1}{2}\left(\frac{\pi}{2}\right)(1) + \frac{1}{4}(1)(0) \right] - \left[ \frac{1}{2}(0)(0) + \frac{1}{4}(0)(1) \right]$$ Simplifying, we get our result: $$= \frac{\pi}{4}$$ So, the definite integral $$\int_{0}^{1} x \sqrt{1-x^{2}} d x$$ is equal to $$\frac{\pi}{4}$$.