Question

In Exercises \(27-40\) , evaluate each integral using Part 2 of the Fundamental Theorem. Support your answer with NINT if you are unsure. $$\int_{0}^{\pi / 3} 2 \sec ^{2} \theta d \theta$$

Short answer

The result is \(2 \sqrt{3}\).

Step-by-step solution

Evaluate Integral

The indefinite integral of \(2 \sec^2 \theta\) is \(2 \tan \theta\), since \(d/d \theta \tan \theta = \sec^2 \theta\).

Fundamental Theorem of Calculus Part 2

The Fundamental Theorem states that the integral of a function from a to b is its antiderivative evaluated at b minus its antiderivative evaluated at a, specifically \(\int_{a}^{b} f(x) dx = F(b) - F(a)\). Here, we substitute the limits of integration \(\theta = 0\) and \(\theta = \pi /3\) into the result from Step 1.

Calculate Result

Substitute \(\pi/3\) and \(0\) into \(2 \tan \theta\). We get \(2 \tan (\pi /3) - 2 \tan (0) = 2 \sqrt{3} - 0 = 2 \sqrt{3}\).