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Chapter 5: Problem 94

Evaluate the following definite integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{\sqrt{3}} \frac{3 d x}{9+x^{2}}$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int_{0}^{\sqrt{3}} \frac{3 dx}{9+x^{2}}$$. Answer: $$\arctan\left(\frac{\sqrt{3}}{3}\right)$$

Step by step solution

01

Identify the given function

The given function is $$\int_{0}^{\sqrt{3}} \frac{3 dx}{9+x^{2}}$$
02

Find the antiderivative of the given function

We first recognize this expression as the derivative of an inverse tangent function. To see this, consider the following: $$\frac{d}{dx}\left(\arctan\left(\frac{x}{3}\right)\right)=\frac{3}{9+x^2}$$ Now, integrate both sides with respect to x: $$\int\frac{d}{dx}\left(\arctan\left(\frac{x}{3}\right)\right)dx=\int\frac{3}{9+x^{2}}dx$$ We know that the antiderivative of the left side is just the function itself: $$\arctan\left(\frac{x}{3}\right) + C = \int\frac{3}{9+x^{2}}dx$$ where C is the constant of integration. However, since we are dealing with a definite integral, we don't need to be concerned about the constant at this moment.
03

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus tells us that to evaluate a definite integral, we need to find the antiderivative of the function and then subtract the antiderivative evaluated at the lower bound of integration from the antiderivative evaluated at the upper bound. So, our definite integral becomes: $$\int_{0}^{\sqrt{3}} \frac{3 dx}{9+x^{2}}=\left[\arctan\left(\frac{x}{3}\right)\right]_{0}^{\sqrt{3}}$$
04

Evaluate the antiderivative at the bounds of integration

Evaluate the antiderivative at the upper and lower bounds and subtract the lower bound expression from the upper bound expression: $$\left[\arctan\left(\frac{\sqrt{3}}{3}\right) - \arctan\left(\frac{0}{3}\right)\right]$$ Now simplify: $$\arctan\left(\frac{\sqrt{3}}{3}\right) - \arctan(0)$$ Since \(\arctan(0)=0\), we have: $$\arctan\left(\frac{\sqrt{3}}{3}\right) - 0 = \arctan\left(\frac{\sqrt{3}}{3}\right)$$
05

Finalize the solution

The definite integral of the given function is: $$\int_{0}^{\sqrt{3}} \frac{3 dx}{9+x^{2}} = \arctan\left(\frac{\sqrt{3}}{3}\right)$$

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