CBSE Class 12 Mathematics Application of Integrals Case Studies

Answer: (b) x² = −250y

Explanation:
1. Take the highest point as the origin; the arch opens downward, so its equation is x² = −4ay.
2. The ends of the arch are 50 ft on either side and 10 ft lower, so (50, −10) lies on it: 2500 = 40a.
3. 4a = 250, so the parabola is x² = −250y.

Teacher's Note:
1. A parabola with its vertex at the top opens downward, so the y-term must carry a minus sign.
2. Use half the span (50 ft), not the full 100 ft, for the x-coordinate of the end point.

Answer: (a) \( \frac{1000}{3} \)

Explanation:
1. The integrand is even, so the integral = \( 2\int_{0}^{50} \frac{x^2}{250}\,dx = \frac{2}{250}\left[\frac{x^3}{3}\right]_0^{50} \)
2. = \( \frac{2}{250} \times \frac{125000}{3} \)
3. = \( \frac{1000}{3} \)

Teacher's Note:
1. \( \int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx \) for an even function halves the work.
2. The answer 0 applies only to odd integrands, not to x².

Answer: (a) Even

Explanation:
1. Let f(x) = x². Then f(−x) = (−x)² = x² = f(x).
2. Since f(−x) = f(x), x² is an even function.

Teacher's Note:
1. Even powers of x give even functions and odd powers give odd functions.
2. Graphically, an even function is symmetric about the y-axis, just like the arch of the bridge.

Answer: (c) \( \frac{1000}{3} \)

Explanation:
1. In the first quadrant x = \( \sqrt{250y} \), so area = \( \int_{0}^{10} \sqrt{250}\,\sqrt{y}\,dy \)
2. = \( \sqrt{250} \cdot \frac{2}{3}\left[y^{3/2}\right]_0^{10} = \frac{2}{3}\sqrt{250} \cdot 10\sqrt{10} \)
3. = \( \frac{2}{3} \times 50 \times 10 = \frac{1000}{3} \) sq. units

Teacher's Note:
1. With the y-axis as boundary, integrate x as a function of y with respect to y.
2. \( \sqrt{250} \times \sqrt{10} = \sqrt{2500} = 50 \) makes the arithmetic neat.

Answer: (d) none of these

Explanation:
1. Area = \( \int_{2}^{4} \sqrt{250y}\,dy = \sqrt{250} \cdot \frac{2}{3}\left[y^{3/2}\right]_2^{4} \)
2. = \( \frac{2}{3} \cdot 5\sqrt{10}\,(8 - 2\sqrt{2}) = \frac{80\sqrt{10} - 40\sqrt{5}}{3} \) ≈ 54.5 sq. units
3. This matches none of (a), (b), (c), so the answer is (d) none of these.

Teacher's Note:
1. Area between y = 2 and y = 4 must be less than the area between y = 0 and y = 10 (1000/3 ≈ 333), which rules out (a) and (c).
2. An area bounded by a curve and lines is never 0 unless the region has no width.