Definite Integration
Grade 12

Question:

<p><span class="math-tex">\(\int\limits_0^{\frac{\pi }{2}} {[\sin x + \cos x]} \)</span>dx, where [ ] represents the greatest integer function, equals</p>
<p style="display:inline"><span class="math-tex">\(\frac {3\pi}{8}\)</span></p>
<p style="display:inline">0</p>
<p style="display:inline"><span class="math-tex">\(\frac {\pi}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac {\pi}{4}\)</span></p>

Step-by-Step Solution

<p>sin x&nbsp;<span class="math-tex">$\geq$</span>&nbsp;sin<sup>2</sup>x and cos x&nbsp;<span class="math-tex">$\geq$</span>&nbsp;cos<sup>2</sup> x in&nbsp;<span class="math-tex">$\left( {0,\frac{\pi }{2}} \right)$</span><br /> <span class="math-tex">$\left.\begin{array}{l} \Rightarrow \sin x+\cos x \geq 1 \\ \text { Also, } \sin x+\cos x \leq \sqrt{2} \end{array}\right\}$</span>&nbsp;<span class="math-tex">$\Rightarrow$</span>&nbsp;[sin x + cos x] = 1<br /> <span class="math-tex">$\int \limits_{0}^{\frac{\pi}{2}}[\sin x+\cos x]$</span>dx =&nbsp;<span class="math-tex">$\int \limits_{0}^{\frac{\pi}{2}} d x=\frac{\pi}{2}$</span></p>
Correct Answer: C

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