Definite Integration
Grade None
Question:
<p>The maximum value of f(x) = <span class="math-tex">\(\int_\limits{0}^{1}\)</span> t sin(x + <span class="math-tex">\(\pi\)</span>t) dt, is</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{\pi} \sqrt{\pi^{2}}+4\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{\pi^{2}} \sqrt{\pi^{2}+4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2 \pi^{2}} \sqrt{\pi^{2}+4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\sqrt{\pi^{2}+4}\)</span></p>
Step-by-Step Solution
Key Concept: Evaluate the integral using integration by parts to express f(x) as a linear combination of sine and cosine, then apply the formula for the maximum value of a trigonometric expression.
<p><span class="math-tex">$f(x) = \int_0^1 {\mathop t\limits_I } \sin \mathop {(\mathop x\limits_{II} + }\limits_{(IBP)} \pi t)dt$</span><br />
<span class="math-tex">$\left.f(x)=\frac{-t}{\pi} \cos (x+\pi t)\right]_{0}^{1}+\frac{1}{\pi} \int_{0}^{1} \cos (x+\pi t) d t$</span><br />
<span class="math-tex">$\left.=\left(\frac{1}{\pi} \cos x-0\right)+\frac{1}{\pi^{2}} \sin (x+\pi t)\right]_{0}^{1}$</span><br />
<span class="math-tex">$f(x)=\frac{1}{\pi} \cos x-\frac{2}{\pi^{2}} \sin x$</span><br />
<span class="math-tex">$\left.\therefore \quad f(x)\right|_{\max }=\sqrt{\frac{1}{\pi^{2}}+\frac{4}{\pi^{4}}}=\frac{\sqrt{\pi^{2}+4}}{\pi^{2}}$</span></p>
Correct Answer: B