MCQ
$\int_0^{\pi /2} {\frac{{\sin x}}{{\sin x + \cos x}}\,dx} $ equals
- A$\frac{\pi }{2}$
- B$\frac{\pi }{3}$
- ✓$\frac{\pi }{4}$
- D$\frac{\pi }{6}$
$\,\,\left( \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right)$
$2I = \int_0^{\pi /2} {dx} \Rightarrow I = \frac{\pi }{4}$.
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$I$. $f$ is continuous on the closed interval $[a, b]$
$II.$ $f$ is bounded on the open interval $(a, b)$
$III.$ If $a$ $< a_1< b_1< b$, and $f (a_1)<0< f (b_1)$, then there is $a$ number $c$ such that $a_1 < c < b_1$ and $f (c)=0$