alleryStart learning →Find the area (in square units) under y = x² from x = 0 to x = 3.
- 27
- 18
- 9
- 3
Show answer and solution
Correct answer: C. 9
The integral of x² from 0 to 3 is 27/3 = 9.
MHT-CET Application of Definite Integration — practice questions
Explore 47 free Application of Definite Integration questions in MHT-CET Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
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Showing questions 1–30 of 47.
Find the area (in square units) under y = x² from x = 0 to x = 3.Read the answer and worked solution →A plot lies under the parabola y = x² for 0 ≤ x ≤ 2; its area isRead the answer and worked solution →Compute the region enclosed by the cubic y = x³, the X-axis and x = 2.Read the answer and worked solution →Area between the line y = 2x, the X-axis and x = 3 (square units) isRead the answer and worked solution →How much area lies beneath the root curve y = √x over [0, 4]?Read the answer and worked solution →Over [0, 3], the curve y = x² + 1 lies above the X-axis; the area below it isRead the answer and worked solution →The curve y = −x² lies below the X-axis; for 1 ≤ x ≤ 4 the area it bounds isRead the answer and worked solution →Splitting at the origin, the area enclosed by y = x, the X-axis, x = −1 and x = 4 isRead the answer and worked solution →One arch of the sine wave y = sin x on [0, π] encloses, with the X-axis, an area ofRead the answer and worked solution →The cosine curve, the X-axis and the Y-axis up to x = π/2 enclose an area ofRead the answer and worked solution →Compute the area below the exponential curve y = eˣ over [0, 1].Read the answer and worked solution →On the reciprocal curve y = 1/x, the strip from x = e to x = e² has areaRead the answer and worked solution →Using logarithms, what is the area beneath y = 1/x between 1 and e³?Read the answer and worked solution →Find the area bounded by y² = 16x and the line x = 4 (square units).Read the answer and worked solution →Find the area bounded by the parabola y² = 8x and its latus rectum.Read the answer and worked solution →Area enclosed by y² = 12x and its latus rectum (square units) isRead the answer and worked solution →Find the area of the region bounded by y² = 9x and x² = 9y.Read the answer and worked solution →The area between the parabolas y² = 4x and x² = 4y isRead the answer and worked solution →The horizontal line y = 4 cuts the parabola y = x²; the enclosed area isRead the answer and worked solution →Area bounded by x² = 16y, y = 1, y = 4 and the Y-axis, in the first quadrant, isRead the answer and worked solution →The area of the circle x² + y² = 25 in the first quadrant isRead the answer and worked solution →What is the area enclosed by the circle x² + y² = 16?Read the answer and worked solution →The area of the ellipse x²/9 + y²/4 = 1 isRead the answer and worked solution →The area under y = 2√x enclosed between x = 0 and x = 1 isRead the answer and worked solution →Between x = 1 and x = 2, the parabola y = x² and the X-axis encloseRead the answer and worked solution →Find the area enclosed between the curves y = x and y = x² (square units).Read the answer and worked solution →The region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 has an area (square units) ofRead the answer and worked solution →The area bounded by y = x³, the X-axis, x = −2 and x = 1 isRead the answer and worked solution →The lines y = x and y = 2x, up to x = 2, enclose a region of areaRead the answer and worked solution →The area enclosed by the parabola y = x² and the line y = 2x isRead the answer and worked solution →
Worked practice questions
A plot lies under the parabola y = x² for 0 ≤ x ≤ 2; its area is
- 4/3
- 8
- 8/3
- 16/3
Show answer and solution
Correct answer: C. 8/3
Area = [x³/3] from 0 to 2 = 8/3.
Compute the region enclosed by the cubic y = x³, the X-axis and x = 2.
- 16
- 2
- 4
- 8
Show answer and solution
Correct answer: C. 4
Area = [x⁴/4] from 0 to 2 = 16/4 = 4.
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