Available Questions 23 found Page 1 of 2
Standalone Questions
#1483
Mathematics
Applications of Integrals
LA
UNDERSTAND
2025
AISSCE(Board Exam)
Competency
5 Marks
A woman discovered a scratch along a straight line on a circular table top of radius 8 cm. She divided the table top into 4 equal quadrants and discovered the scratch passing through the origin inclined at an angle $\frac{\pi}{4}$ anticlockwise along the positive direction of x-axis. Find the area of the region enclosed by the x-axis, the scratch and the circular table top in the first quadrant, using integration.
Key:
Sol:
Sol:
#1461
Mathematics
Applications of Integrals
LA
REMEMBER
2025
AISSCE(Board Exam)
Competency
5 Marks
Draw a rough sketch for the curve $y=2+|x+1|$. Using integration, find the area of the region bounded by the curve $y=2+|x+1|$, $x=-4$, $x=3$ and $y=0$.
Key:
Sol:
Sol:
#1437
Mathematics
Applications of Integrals
LA
UNDERSTAND
2025
AISSCE(Board Exam)
Competency
5 Marks
Sketch a graph of $y=x^{2}$. Using integration, find the area of the region bounded by $y=9$, $x=0$ and $y=x^{2}$.
Key:
Sol:
Sol:
#1404
Mathematics
Applications of Integrals
VSA
UNDERSTAND
2025
AISSCE(Board Exam)
KNOWLEDGE
2 Marks
Calculate the area of the region bounded by the curve $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the x-axis using integration.
Key:
Sol:
Sol:
#1393
Mathematics
Applications of Integrals
LA
REMEMBER
2025
AISSCE(Board Exam)
KNOWLEDGE
5 Marks
Using integration, find the area of the region bounded by the line $y=5x+2$, the x-axis and the ordinates $x=-2$ and $x=2$.
Key:
Sol:
Sol:
#1370
Mathematics
Applications of Integrals
SA
UNDERSTAND
2025
AISSCE(Board Exam)
Competency
3 Marks
Sketch the graph of $y=|x+3|$ and find the area of the region enclosed by the curve, x-axis, between $x=-6$ and $x=0$, using integration.
Key:
Sol:
Sol:
#1351
Mathematics
Applications of Integrals
LA
UNDERSTAND
2024
AISSCE(Board Exam)
KNOWLEDGE
5 Marks
Find the area of the region bounded by the curve $4x^{2}+y^{2}=36$ using integration.
Key:
Sol:
Sol:
#1327
Mathematics
Applications of Integrals
LA
UNDERSTAND
2024
AISSCE(Board Exam)
KNOWLEDGE
5 Marks
Using integration, find the area bounded by the ellipse $9x^{2}+25y^{2}=225$, the lines $x=-2,$ $x=2$, and the X-axis.
Key:
Sol:
Sol:
#1326
Mathematics
Applications of Integrals
LA
REMEMBER
2024
AISSCE(Board Exam)
Competency
5 Marks
Sketch the graph of $y=x|x|$ and hence find the area bounded by this curve, X-axis and the ordinates $x=-2$ and $x=2,$ using integration.
Key:
Sol:
Sol:
#1308
Mathematics
Applications of Integrals
LA
REMEMBER
2024
AISSCE(Board Exam)
KNOWLEDGE
5 Marks
Using integration, find the area of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1,$ included between the lines $x=-2$ and $x=2$.
Key:
Sol:
Sol:
#1287
Mathematics
Applications of Integrals
LA
REMEMBER
2024
AISSCE(Board Exam)
KNOWLEDGE
5 Marks
If $A_{1}$ denotes the area of region bounded by $y^{2}=4x,$ $x=1$ and x-axis in the first quadrant and $A_{2}$ denotes the area of region bounded by $y^{2}=4x,$ $x=4$, find $A_{1}:A_{2}$.
Key:
Sol:
Sol:
#1263
Mathematics
Applications of Integrals
LA
REMEMBER
2024
AISSCE(Board Exam)
KNOWLEDGE
5 Marks
Using integration, find the area of the region enclosed between the circle $x^{2}+y^{2}=16$ and the lines $x=-2$ and $x=2.$
Key:
Sol:
Sol:
#930
Mathematics
Applications of Integrals
LA
APPLY
2023
Competency
5 Marks
The area of the region bounded by the line \(y=mx (m>0)\), the curve \(x^{2}+y^{2}=4\) and the \(x\)-axis in the first quadrant is \(\frac{\pi}{2}\) units. Using integration, find the value of m.
Key:
Sol:
Sol:
#929
Mathematics
Applications of Integrals
LA
APPLY
2023
Competency
5 Marks
33. Using integration, find the area of the region bounded by the parabola $y^{2}=4ax$ and its latus rectum.
Key:
Sol:
Sol:
#874
Mathematics
Applications of Integrals
LA
APPLY
2023
Competency
5 Marks
Find the area of the region bounded by the curves x^{2}=y, y=x+2 and x-axis, using integration.
Key:
Sol:
Sol:
#873
Mathematics
Applications of Integrals
SA
APPLY
2023
Competency
3 Marks
Find the area of the following region using integration: {(x,y): y² ≤ 2x and y ≥ x-4}
Key:
Sol:
Sol:
#872
Mathematics
Applications of Integrals
VSA
APPLY
2023
Competency
2 Marks
Sketch the region bounded by the lines 2x+y=8, y=2, y=4 and the y-axis. Hence, obtain its area using integration.
Key:
Sol:
Sol:
#642
Mathematics
Applications of Integrals
MCQ_SINGLE
APPLY
2025
AISSCE(Board Exam)
KNOWLEDGE
1 Marks
The area of the shaded region (figure) represented by the curves \(y=x^{2}\), \(0\le x\le2\) and y-axis is given by(A) \(\int_{0}^{2} x^2 dx\)
(B) \(\int_{0}^{2} \sqrt{y} dy\)
(C) \(\int_{0}^{4} x^2 dx\)
(D) \(\int_{0}^{4} \sqrt{y} dy\)
Key: D
Sol:
Sol:
\(\int_{0}^{4} \sqrt{y} dy\)
#641
Mathematics
Applications of Integrals
MCQ_SINGLE
APPLY
2025
AISSCE(Board Exam)
KNOWLEDGE
1 Marks
The area of the region enclosed by the curve \(y=\sqrt{x}\) and the lines \(x=0\) and \(x=4\) and x-axis is:
(A) \(\frac{16}{9}\) sq. units
(B) \(\frac{32}{9}\) sq. units
(C) \(\frac{16}{3}\) sq. units
(D) \(\frac{32}{3}\) sq. units
Key: C
Sol:
Sol:
The area (\(A\)) of the region enclosed by the curve \(y = f(x)\), the \(x\)-axis, and the vertical lines \(x=a\) and \(x=b\) is given by the definite integral:\[A = \int_{a}^{b} f(x) dx\]
\[A = \int_{0}^{4} x^{1/2} dx\]
\[A = \frac{2}{3} \left[ x^{3/2} \right]_{0}^{4}\]
\[A = \frac{2}{3} \left[ (4)^{3/2} - (0)^{3/2} \right]\]
\[A = \frac{16}{3}\]
#640
Mathematics
Applications of Integrals
MCQ_SINGLE
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
The area of the region enclosed between the curve \(y=x|x|\), x-axis, \(x=-2\) and \(x=2\) is:
(A) \(\frac{8}{3}\)
(B) \(\frac{16}{3}\)
(C) 0
(D) 8
Key:
Sol:
Sol:
The function is given by $y=x|x|$. We can define this piecewise:
$$y = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}$$
The area $A$ enclosed between the curve, the x-axis, $x=-2$ and $x=2$ is given by the integral of the absolute value of the function over the interval $[-2, 2]$:
$$A = \int_{-2}^{2} |y| \, dx = \int_{-2}^{2} |x|x|| \, dx$$
Since $|x| \ge 0$, we have $|x|x|| = |x||x| = x^2$.
So, the integral becomes:
$$A = \int_{-2}^{2} x^2 \, dx$$
We can evaluate this integral:
$$A = \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{(2)^3}{3} - \frac{(-2)^3}{3}$$
$$A = \frac{8}{3} - \frac{-8}{3} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}$$
The final answer is $\boxed{\frac{16}{3}}$.