IMO Practice Test — Application of Integrals
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
medium
Area between $y=x^2$ and $y=\sqrt{x}$ is:
$1/3$
$2/3$
$1/6$
$1$
Explanation: Intersections: $(0,0)$ and $(1,1)$. $A=\int_0^1(\sqrt{x}-x^2)dx=[2x^{3/2}/3-x^3/3]_0^1=2/3-1/3=1/3$.
Question 2 of 6
hard
The area of the region $\{(x,y): x^2+y^2\le 1, x+y\ge0\}$ is:
$\pi/2$
$\pi/4$
$\pi$
$\pi/3$
Explanation: The region is the half of the unit circle where $x+y\ge0$ (the half above the line $y=-x$). This is exactly half the circle, so area $=\pi(1)^2/2=\pi/2$.
Question 3 of 6
medium
Area between the parabolas $y^2=x$ and $x^2=y$ is:
$1/3$
$2/3$
$1/6$
$1/4$
Explanation: Both intersect at $(0,0)$ and $(1,1)$. Upper curve $y=\sqrt{x}\ge x^2$ on $[0,1]$. $A=\int_0^1(\sqrt{x}-x^2)dx=1/3$.
Question 4 of 6
hard
The area bounded by $y=|x-1|$ and $y=5-x^2$ is:
$\frac{31}{3}$
$\frac{22}{3}$
$\frac{32}{3}$
$\frac{20}{3}$
Explanation: Careful analysis of intersections for $x<1$ and $x>1$ cases, then integrate.
Question 5 of 6
hard
Ratio of areas cut off from $x^2+y^2=4$ by the line $y=1$:
$\frac{2\pi-3\sqrt{3}}{4\pi+3\sqrt{3}}$
$\frac{\pi-3\sqrt{3}}{3\pi}$
$(4\pi-3\sqrt{3}):(8\pi+3\sqrt{3})$
$(4\pi-3\sqrt{3}):(4\pi+3\sqrt{3})$
Explanation: Area above $y=1$: integrate $\sqrt{4-y^2}$ from 1 to 2, double. Area below: subtract from $4\pi$.
Question 6 of 6
hard
Area of the region bounded by $y=x$, $y=x^3$ for $-1\le x\le 1$ is:
$1/2$
$0$
$1$
$1/4$
Explanation: On $[0,1]$: $x\ge x^3$. Area $=2\int_0^1(x-x^3)dx=2[x^2/2-x^4/4]_0^1=2(1/4)=1/2$.