Online Test — Application of Integrals
10 Questions • 20 min • Chapter MCQ
20:00
Question 1 of 10
easy
Area bounded by $y=x^2$ and $y=x$ is:
$1/6$
$1/3$
$1/2$
$1$
Explanation: Intersections: $x^2=x \Rightarrow x=0,1$. $A=\int_0^1(x-x^2)dx=[x^2/2-x^3/3]_0^1=1/2-1/3=1/6$.
Question 2 of 10
easy
Area of the circle $x^2+y^2=4$ is:
$4\pi$
$2\pi$
$\pi$
$8\pi$
Explanation: Circle with $r=2$: area $=\pi r^2=4\pi$.
Question 3 of 10
easy
Area bounded by $y=\sin x$ and $x$-axis from $0$ to $\pi$ is:
$0$
$1$
$2$
$\pi$
Explanation: $\int_0^\pi\sin x\,dx=[-\cos x]_0^\pi=1-(-1)=2$.
Question 4 of 10
medium
The area between $y=4x-x^2$ and $y=x$ is:
$9/2$
$4$
$9$
$3/2$
Explanation: $4x-x^2=x\Rightarrow 3x=x^2\Rightarrow x=0,3$. $A=\int_0^3(4x-x^2-x)dx=\int_0^3(3x-x^2)dx=[3x^2/2-x^3/3]_0^3=27/2-9=9/2$.
Question 5 of 10
medium
The area enclosed by $|x|+|y|=1$ is:
$1$
$2$
$4$
$1/2$
Explanation: The region is a square with diagonals of length 2. Area of square with diagonal $d=2$: $d^2/2=2$.
Question 6 of 10
easy
Area of ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is:
$12\pi$
$9\pi$
$16\pi$
$25\pi$
Explanation: $a=4$, $b=3$. Area $=\pi ab=12\pi$.
Question 7 of 10
medium
Area bounded by $y^2=x$ and $x=4$ is:
$16/3$
$8/3$
$32/3$
$4$
Explanation: $A=2\int_0^4\sqrt{x}\,dx=2[\frac{2}{3}x^{3/2}]_0^4=\frac{4}{3}(8)=32/3$.
Question 8 of 10
medium
Area between $y=\cos x$ and $y=\sin x$ for $x\in[0,\pi/4]$ is:
$\sqrt{2}-1$
$\sqrt{2}+1$
$2-\sqrt{2}$
$\sqrt{2}$
Explanation: On $[0,\pi/4]$: $\cos x\ge\sin x$. $A=\int_0^{\pi/4}(\cos x-\sin x)dx=[\sin x+\cos x]_0^{\pi/4}=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})-(0+1)=\sqrt{2}-1$.
Question 9 of 10
medium
The area of the region bounded by $y=x^3$, $x$-axis, $x=1$ and $x=-1$ is:
$0$
$1/2$
$1/4$
$1$
Explanation: $y=x^3$ is odd. Area (taking absolute values) $=2\int_0^1 x^3 dx=2[x^4/4]_0^1=1/2$.
Question 10 of 10
easy
Area bounded by $x^2=y$ and $x=y$ is:
$1/6$
$1/3$
$1/2$
$1$
Explanation: $x^2=x\Rightarrow x=0,1$. $A=\int_0^1(x-x^2)dx=1/2-1/3=1/6$.