InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Find the area bounded by the curve `y=(4-x^2)` the y-axis and the lines y=0 y=3 |
| Answer» Required area `=underset(-2)overset(3)intx dy = underset(0)overset(3)int sqrt(4-y ) dy ` | |
| 2. |
Find the area of the region in the first quadrant enclosed by the x-axis, the line `y = x`, and the circle `x^2+y^2=32`. |
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Answer» Required area `=underset(4)overset(0)int xdx+underset(4)overset(4sqrt(2))int sqrt(32-x^2)dx` `=[x^2/2]_0^4+[(xsqrt(32-x^2))/2+32/2sin^-1""x/(4sqrt(2))]_4^(4sqrt(2))` `4 pi` sq units |
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| 3. |
Find the area of the region bounded by the curve `y^2=4x`and the line `x = 3`. |
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Answer» Since the equation `y^2=4x` contains only even powers of y the curve is symmetrical about the y - axis `therefore ` required area = `2.underset(0)overset(3)int2sqrt(x)`dx |
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| 4. |
Using integration, find the area of the region bounded by the parabola `y^2=16x` and the line `x=4` |
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Answer» `y^2=16x` is a right-handed parabola with its vertix at the origin And x=4 is the line parallel to the y-axis at a distance of 4 units from it . |
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