MCQ
Area between the curve $y = \cos x$ and $x - $ axis when $0 \le x \le 2\pi $ is
- A$2$
- ✓$4$
- C$0$
- D$3$
✓
Answer
Correct option: B.
$4$
b
(b) $y = \cos x$,
(b) $y = \cos x$,
When $x \in \left[ {0,\frac{\pi }{2}} \right],\cos x \ge 0$
When $x \in \left[ {\frac{\pi }{2},\frac{{3\pi }}{2}} \right],\cos x \le 0$
When $x \in \left[ {\frac{{3\pi }}{2},2\pi } \right],\cos x \ge 0$
Thus required area is given by,
$\int_0^{\pi /2} {\,\,ydx} = \int_0^{\pi /2} {\cos x\,dx + \int_{\pi /2}^{3\pi /2} {( - \cos x)dx + \int_{3\pi /2}^{2\pi } {\,\cos xdx} } } $
$ = 1 + 2 + 1 = 4\,\, sq. \,unit$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The feasible region for an $\text{LPP}$ is shown below:
Let $Z = 3x - 4y$ be the objective function. Minimum of $Z$ occurs at
→Let $Z = 3x - 4y$ be the objective function. Minimum of $Z$ occurs at
For $\text{x, }\in\text{ R},\text{f}(\text{x})=\mid\log2-\sin\text{x}\mid$ and $g(x) = f(f(x))$ then :
→If $\left(\frac{1}{2}, \frac{1}{3}, n\right)$ are the direction cosines of a line, then the value of $n$ is
→$\int {\frac{{\sec x\;dx}}{{\sqrt {\cos 2x} }}} = $
→Function $\text{f}(\text{x})=\log_\text{a}\text{x}$ is increasing on $R,$ if:
→z = 10x + 25y subject to $0\leq\text{X}\leq3$ and $0\leq\text{X}\leq3,$ $\text{x}+\text{y}\leq5$ then the maximum value of z is:
→The order and degree of the differential equation $\Big(1+3\frac{\text{dy}}{\text{dx}}\Big)^\frac{2}{3}=4\frac{\text{d}^3\text{y}}{\text{dx}^3}$ are:
→The scalar product of $5i + j - 3k$ and $3i - 4j + 7k$ is:
→A root of the equation $\left| {\,\begin{array}{*{20}{c}}{3 - x}&{ - 6}&3\\{ - 6}&{3 - x}&3\\3&3&{ - 6 - x}\end{array}\,} \right| = 0$ is
→Let $u,\,v,\,w$ be such that $|u|\, = 1,\,|v|\, = 2,\,|w|\, = 3.$ If the projection $v$ along $u$ is equal to that of $w$ along $u$ and $v,\,\,w$ are perpendicular to each other then $|u - v + w|$ equals
→