Question

a graphing calculator is recommended.
what is wrong with the equation?
\\( \int_{\pi / 3}^{\pi} 9 \sec (\theta) \tan (\theta) d \theta=9 \sec (\theta) _{\pi / 3}^{\pi}=-27 \\)
\\( \bigcirc \\) the lower limit is not equal to 0, so part 2 of the fundamental theorem of calculus cannot be applied.
\\( \bigcirc f(\theta)=9 \sec (\theta) \\) is not continuous at \\( \theta=\pi / 3 \\) so part 2 of the fundamental theorem of calculus cannot be applied.
\\( \bigcirc \\) there is nothing wrong with the equation.
\\( \bigcirc f(\theta)=9 \tan (\theta) \\) is not continuous on the interval \\( \pi / 3, \pi \\) so part 2 of the fundamental theorem of calculus cannot be applied.
\\( \bigcirc f(\theta)=9 \sec (\theta) \tan (\theta) \\) is not continuous on the interval \\( \pi / 3, \pi \\) so part 2 of the fundamental theorem of calculus cannot be applied.
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Explanation:

To apply the second part of the Fundamental Theorem of Calculus (\(\int_{a}^{b}f(x)dx = F(b)-F(a)\) where \(F^\prime(x)=f(x)\)), the function \(f(x)\) must be continuous on the interval \([a, b]\).

The function \(y = \sec\theta=\frac{1}{\cos\theta}\) has vertical asymptotes where \(\cos\theta = 0\). The function \(y=\tan\theta=\frac{\sin\theta}{\cos\theta}\) also has vertical asymptotes where \(\cos\theta = 0\).

For the function \(f(\theta)=9\sec\theta\tan\theta=\frac{9\sin\theta}{\cos^{2}\theta}\), \(\cos\theta = 0\) when \(\theta=\frac{\pi}{2}\in[\frac{\pi}{3},\pi]\). At \(\theta=\frac{\pi}{2}\), \(\lim_{\theta
ightarrow\frac{\pi}{2}^{-}}\frac{9\sin\theta}{\cos^{2}\theta}=\infty\) and \(\lim_{\theta
ightarrow\frac{\pi}{2}^{+}}\frac{9\sin\theta}{\cos^{2}\theta}=-\infty\). So \(f(\theta) = 9\sec\theta\tan\theta\) is not continuous on \([\frac{\pi}{3},\pi]\)

Let's check other options:

• The Fundamental Theorem of Calculus (Part 2) does not require the lower - limit \(a = 0\). The formula is \(\int_{a}^{b}f(x)dx=F(b)-F(a)\) for any \(a,b\) as long as \(f\) is continuous on \([a,b]\).
• \(F(\theta)=9\sec\theta\) is differentiable (and thus continuous) on intervals where \(\cos\theta

eq0\). The problem is with the integrand \(f(\theta)=9\sec\theta\tan\theta\), not with \(F(\theta)\).

• \(f(\theta) = 9\tan\theta\) is not the integrand. The integrand is \(f(\theta)=9\sec\theta\tan\theta\)

Answer:

\(f(\theta)=9\sec(\theta)\tan(\theta)\) is not continuous on the interval \([\pi/3,\pi]\) so part 2 of the fundamental theorem of calculus cannot be applied.