JEE Mathematics Area Under The Curve MCQs Set 06

Answer: (c) \( \int_{\alpha}^{\beta} f(x)\,dx \)

Explanation:
1. The ordinates through P and Q are the lines x = \( \alpha \) and x = \( \beta \).
2. The area under the curve y = f(x) between these lines and above the x-axis is \( \int_{\alpha}^{\beta} f(x)\,dx \).

Teacher's Note:
1. Area under a curve is always integrated along the axis on which the boundary lines are marked.
2. The limits must be the values of the variable of integration, here x = \( \alpha \) to x = \( \beta \), not the function values.

Answer: (a) 13 unit²

Explanation:
1. The boundaries are x = −3, x = 3, the x-axis and the V-shaped graph of y = |x − 2|.
2. For −3 ≤ x ≤ 2, y = 2 − x, so the area is \( \int_{-3}^{2}(2-x)\,dx = 12.5 \).
3. For 2 ≤ x ≤ 3, y = x − 2, so the area is \( \int_{2}^{3}(x-2)\,dx = 0.5 \).
4. Total area = 12.5 + 0.5 = 13 unit².

Teacher's Note:
1. Split the integral wherever the expression inside a modulus changes sign.
2. Here the break point is x = 2.

Answer: (b) \( 2\pi \)

Explanation:
1. Squaring gives x² + y² = 4 with y ≥ 0, which is the upper semicircle of radius 2.
2. Area = ½ × π × 2² = 2π.

Teacher's Note:
1. Recognise standard curves before integrating; a semicircle needs no calculus.
2. Remember y = +√ means only the upper half of the circle.

Answer: (c) \( \frac{16}{3}\sqrt{2} \) unit²

Explanation:
1. By symmetry about the x-axis, Area = \( 2\int_0^2 2\sqrt{x}\,dx \).
2. = \( 4\cdot\frac{2}{3}x^{3/2}\Big|_0^2 = \frac{8}{3}\cdot 2\sqrt{2} \).
3. = \( \frac{16}{3}\sqrt{2} \) unit².

Teacher's Note:
1. A double ordinate is a chord perpendicular to the axis of the parabola, so the area is twice the upper half.
2. Use \( \int x^{1/2}dx = \frac{2}{3}x^{3/2} \).

Answer: (a) 3

Explanation:
1. Area = \( 2\int_0^3 \sqrt{ky}\,dy = 2\sqrt{k}\cdot\frac{2}{3}\cdot 3\sqrt{3} = 4\sqrt{3k} \).
2. Setting \( 4\sqrt{3k} = 12 \) gives \( \sqrt{3k}=3 \), so 3k = 9 and k = 3.

Teacher's Note:
1. Integrating with respect to y is easier here because the line is y = 3.
2. Check by substitution: x² = 3y and y = 3 meet at x = ±3, area = \( \int_{-3}^{3}(3 - x^2/3)dx = 12 \).

Answer: (d) \( \log_2 e \)

Explanation:
1. The curve lies above the x-axis and approaches it as x → −∞, so the bounded region is for x from −∞ to 0.
2. Area = \( \int_{-\infty}^{0} 2^x dx = \frac{2^x}{\ln 2}\Big|_{-\infty}^{0} = \frac{1}{\ln 2} \).
3. \( \frac{1}{\ln 2} = \log_2 e \).

Teacher's Note:
1. An unbounded region can still have finite area when the curve approaches the axis fast enough.
2. Remember \( 1/\ln a = \log_a e \).

Answer: (a) \( \frac{a^2}{6} \)

Explanation:
1. y = (√a − √x)² = a − 2√(a x) + x for 0 ≤ x ≤ a.
2. Area = \( \int_0^a (a - 2\sqrt{a}\,x^{1/2} + x)\,dx = a^2 - \frac{4}{3}a^2 + \frac{a^2}{2} \).
3. = \( \frac{a^2}{6} \).

Teacher's Note:
1. Expand the square carefully before integrating.
2. The curve is a part of a parabola touching both axes at (a, 0) and (0, a).

Answer: (c) 2 sin 1

Explanation:
1. On [−1, 1] we have cos x > 0 since 1 radian is less than π/2.
2. Area = \( \int_{-1}^{1}\cos x\,dx = 2\sin 1 \).

Teacher's Note:
1. For an even function, integrate over half the interval and double.
2. Always check that the function does not change sign in the interval.

Answer: (a) 9

Explanation:
1. Solving \( \sqrt{x} \) = (x − 3)/2 gives x = 9 and y = 3; the line meets the x-axis at x = 3.
2. Area under the curve from 0 to 9 = \( \frac{2}{3}\cdot 27 = 18 \).
3. Subtract the triangle under the line between x = 3 and x = 9: ½ × 6 × 3 = 9.
4. Required area = 18 − 9 = 9.

Teacher's Note:
1. When two curves enclose a region with the axis, subtract the area under the lower boundary from that under the upper one.
2. A rough sketch avoids mistakes about which region is meant.

Answer: (b) 4 unit²

Explanation:
1. The graphs meet at (−1, 2) and (2, 1); between x = −1 and x = 2 the graph of 3 − |x| lies above |x − 1|.
2. Area = \( \int_{-1}^{2}(3-|x|-|x-1|)\,dx \).
3. Pieces: 1 on [−1, 0], 2 on [0, 1], 1 on [1, 2].
4. Total = 1 + 2 + 1 = 4 unit².

Teacher's Note:
1. Break the integrand at x = 0 and x = 1 where the moduli change.
2. The region is a quadrilateral, so the answer can be cross-checked with plane geometry.

Answer: (c) 15 : 49

Explanation:
1. The curves meet at (0, 0) and (4, 4); total area = \( \int_0^4(2\sqrt{x}-x^2/4)dx = 16/3 \).
2. Part for 0 ≤ x ≤ 1 is \( \frac{4}{3} - \frac{1}{12} = \frac{5}{4} \).
3. The other part is \( \frac{16}{3}-\frac{5}{4}=\frac{49}{12} \).
4. Ratio = 5/4 : 49/12 = 15 : 49.

Teacher's Note:
1. Compute the total area first, then one part, and obtain the other by subtraction.
2. Do not forget to reduce the ratio.

Answer: (a) \( \frac{1}{2} \)

Explanation:
1. Since \( \int_0^a f(x)dx \) equals the given expression, differentiate with respect to a.
2. f(a) = a + ½ sin a + (a/2) cos a − (π/2) sin a.
3. At a = π/2: f = π/2 + ½ + 0 − π/2 = ½.

Teacher's Note:
1. Differentiating an area function with respect to its upper limit gives back the integrand.
2. Apply the product rule to the term (a/2) sin a.

Answer: (b) \( |f(p)-f(q)| \)

Explanation:
1. Area = \( \int_p^q c e^x dx = c(e^q - e^p) \).
2. Since f(q) − f(p) = c(e^q − e^p), the area equals |f(p) − f(q)|, so it is proportional to |f(p) − f(q)|.

Teacher's Note:
1. The exponential function is its own antiderivative, so area equals the difference of ordinates.
2. The modulus takes care of p > q.

Answer: (c) \( 2\sqrt{2} \)

Explanation:
1. Total area = \( \int_2^4 (1+8/x^2)dx = 2 + 8(\tfrac12-\tfrac14) = 4 \), so each part is 2.
2. \( \int_2^a(1+8/x^2)dx = (a-2)+8(\tfrac12-\tfrac1a) = 2 \).
3. This gives a + 2 − 8/a = 2, so a² = 8 and a = 2√2 (it lies between 2 and 4).

Teacher's Note:
1. Use half of the total area as the target for the partial integral.
2. Reject roots outside the given interval.

Answer: (b) \( \frac{1}{12}(4\pi-3\sqrt{3}) \)

Explanation:
1. The ellipse has centre (−1, 0), semi-axes 2 and 1, so in the first quadrant x runs from 0 to 1 with \( y=\sqrt{1-(x+1)^2/4} \).
2. Put u = (x + 1)/2: area = \( 2\int_{1/2}^{1}\sqrt{1-u^2}\,du = \left[u\sqrt{1-u^2}+\sin^{-1}u\right]_{1/2}^{1} \).
3. = π/2 − (√3/4 + π/6) = π/3 − √3/4 = (4π − 3√3)/12.

Teacher's Note:
1. The ellipse is shifted left, so only the part with x between 0 and 1 is in the first quadrant.
2. Use the standard integral \( \int\sqrt{1-u^2}du = \frac12(u\sqrt{1-u^2}+\sin^{-1}u) \).