JEE Mathematics Area Under The Curve MCQs Set 09

Answer: (c) \( 4(4-\sqrt2) \) sq units

Explanation:
1. Area \( =\int_2^4 3\sqrt x\,dx=2\left[x^{3/2}\right]_2^4=2(8-2\sqrt2) \).
2. \( =16-4\sqrt2=4(4-\sqrt2) \) sq units.

Teacher's Note:
1. Integrate \( y=3\sqrt x \) between the given ordinates.

Answer: (b) \( 8\sqrt3 \) sq units

Explanation:
1. By symmetry about the x-axis, area \( =2\int_0^3 2\sqrt x\,dx=4\cdot\frac23\cdot3^{3/2} \).
2. \( =\frac83\cdot3\sqrt3=8\sqrt3 \) sq units.

Teacher's Note:
1. Double the area above the axis.

Answer: (a) \( \pi \) sq units

Explanation:
1. For \( x=0 \) to \( x=2 \) the first quadrant part of the circle is a quarter of the whole circle.
2. Area \( =\frac14\pi(2)^2=\pi \) sq units.

Teacher's Note:
1. The region is exactly a quarter of the circle.

Answer: (a) 9 sq units

Explanation:
1. The graph is a V with vertex at \( (-3,0) \); from \( -6 \) to \( -3 \) and \( -3 \) to 0 there are two triangles with base 3 and height 3.
2. Area \( =2\cdot\frac12\cdot3\cdot3=9 \).

Teacher's Note:
1. Split the interval at the vertex \( x=-3 \).

Answer: (c) \( \frac23 \)

Explanation:
1. The curve is odd, positive for \( x>0 \) and negative for \( x<0 \), so the areas on both sides add.
2. Area \( =2\int_0^1x^2\,dx=\frac23 \).

Teacher's Note:
1. Do not let the two halves cancel; areas are positive.

Answer: (a) 96 sq units

Explanation:
1. Area \( =\int_2^8\frac{5x+7}{2}\,dx=\left[\frac{5x^2}{4}+\frac{7x}2\right]_2^8 \).
2. \( =(80+28)-(5+7)=96 \) sq units.

Teacher's Note:
1. The line lies above the x-axis for \( x\in[2,8] \).

Answer: (c) \( \frac{16}3 \) sq units

Explanation:
1. Area \( =\int_1^5\sqrt{x-1}\,dx=\frac23(x-1)^{3/2}\Big|_1^5 \).
2. \( =\frac23\cdot8=\frac{16}3 \).

Teacher's Note:
1. Substitute \( u=x-1 \).

Answer: (a) \( \frac35a^2\left(2\cdot2^{2/3}-1\right) \) sq unit

Explanation:
1. Express x in terms of y: \( x=a^{1/3}y^{2/3} \), and integrate along the y-axis.
2. Area \( =\int_a^{2a}a^{1/3}y^{2/3}\,dy=a^{1/3}\cdot\frac35\left[(2a)^{5/3}-a^{5/3}\right]=\frac35a^2\left(2\cdot2^{2/3}-1\right) \).

Teacher's Note:
1. When the boundary lines are horizontal, integrate with respect to y.

Answer: (c) \( \frac92 \) sq units

Explanation:
1. The curve meets the y-axis where \( 2-y-y^2=0 \), i.e. \( y=-2 \) and \( y=1 \).
2. Area \( =\int_{-2}^1(2-y-y^2)\,dy=\left[2y-\frac{y^2}2-\frac{y^3}3\right]_{-2}^1=\frac76-\left(-\frac{10}{3}\right)=\frac92 \).

Teacher's Note:
1. Find the limits from the y-intercepts.

Answer: (a) 1 sq unit

Explanation:
1. The curve is \( y=1-|x| \), a triangle with vertices \( (-1,0) \), \( (1,0) \) and \( (0,1) \).
2. Area \( =\frac12\cdot2\cdot1=1 \) sq unit.

Teacher's Note:
1. The graph is a tent over the x-axis.

Answer: (b) \( 3+16\log2 \) sq units

Explanation:
1. \( y=\frac{3x+10}{x-2}=3+\frac{16}{x-2} \), which is positive on \( [3,4] \).
2. Area \( =\int_3^4\left(3+\frac{16}{x-2}\right)dx=3+16\ln2 \) sq units.

Teacher's Note:
1. Divide to separate the constant part.

Answer: (b) \( \pi \) sq units

Explanation:
1. Area \( =\int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\left[\tan^{-1}x\right]_{-\infty}^{\infty} \).
2. \( =\frac\pi2+\frac\pi2=\pi \).

Teacher's Note:
1. Use the standard integral of \( \frac1{1+x^2} \).

Answer: (a) \( \log2 \) sq unit

Explanation:
1. \( y=\frac1x \), so area \( =\int_1^2\frac{dx}{x}=\ln2 \).

Teacher's Note:
1. Integrate \( 1/x \) from 1 to 2.

Answer: (a) \( \frac{125}6 \) sq units

Explanation:
1. The roots of \( 4+3x-x^2=0 \) are \( x=-1 \) and \( x=4 \).
2. Area \( =\int_{-1}^4(4+3x-x^2)\,dx=\left[4x+\frac{3x^2}2-\frac{x^3}3\right]_{-1}^4=\frac{125}6 \).

Teacher's Note:
1. The area of a parabola segment is \( \frac{|a|(\text{root diff})^3}{6} \).

Answer: (b) 1

Explanation:
1. Area \( =\int_0^22^{kx}dx=\frac{2^{2k}-1}{k\ln2} \).
2. Setting this equal to \( \frac3{\ln2} \): \( \frac{4^k-1}{k}=3 \), which holds for \( k=1 \).

Teacher's Note:
1. Test the options in \( \frac{4^k-1}k=3 \).

Answer: (d) \( \left(\frac\pi{4\sqrt2}+\frac1{\sqrt2}-1\right) \) sq unit

Explanation:
1. Area \( =\int_0^{1/\sqrt2}\sin^{-1}x\,dx=\left[x\sin^{-1}x+\sqrt{1-x^2}\right]_0^{1/\sqrt2} \).
2. \( =\frac1{\sqrt2}\cdot\frac\pi4+\frac1{\sqrt2}-1 \).

Teacher's Note:
1. Integrate by parts, or use the formula for \( \int\sin^{-1}x\,dx \).

Answer: (b) \( \left(\frac\pi4-\log\sqrt2\right) \) sq unit

Explanation:
1. Area \( =\int_0^1\tan^{-1}x\,dx=\left[x\tan^{-1}x-\frac12\ln(1+x^2)\right]_0^1 \).
2. \( =\frac\pi4-\frac12\ln2=\frac\pi4-\log\sqrt2 \).

Teacher's Note:
1. Integrate by parts.

Answer: (b) \( \frac\pi4 \) sq unit

Explanation:
1. \( \int_{\pi/2}^{\pi}\sin^2x\,dx=\int_{\pi/2}^\pi\frac{1-\cos2x}2dx=\frac12\cdot\frac\pi2-\frac14[\sin2x]_{\pi/2}^\pi \).
2. \( =\frac\pi4-0=\frac\pi4 \).

Teacher's Note:
1. Use \( \sin^2x=\frac{1-\cos2x}2 \).

Answer: (c) \( 2\sqrt3 \) sq units

Explanation:
1. Area \( =\int_{-\pi/3}^{\pi/3}\sec^2x\,dx=[\tan x]_{-\pi/3}^{\pi/3} \).
2. \( =\sqrt3+\sqrt3=2\sqrt3 \).

Teacher's Note:
1. The integral of \( \sec^2x \) is \( \tan x \).

Answer: (a) 1

Explanation:
1. Area \( =\int_0^\infty e^{-x}dx=\left[-e^{-x}\right]_0^\infty=1 \).

Teacher's Note:
1. The improper integral converges to 1.

Answer: (c) 8

Explanation:
1. \( x^3-4x=x(x-2)(x+2) \), with roots \( -2,0,2 \). The curve is odd, so the two loops have equal area.
2. Area \( =2\int_0^2(4x-x^3)\,dx=2\left[2x^2-\frac{x^4}4\right]_0^2=2(8-4)=8 \).

Teacher's Note:
1. Use symmetry of the odd function.

Answer: (b) 1 sq unit

Explanation:
1. \( \ln x \) is non-negative on \( [1,e] \).
2. Area \( =\int_1^e\ln x\,dx=[x\ln x-x]_1^e=(e-e)-(0-1)=1 \).

Teacher's Note:
1. The curve meets the x-axis at \( x=1 \).

Answer: (b) \( \frac{a^2}2\left(\frac\pi2-1\right) \) sq units

Explanation:
1. The chord subtends \( 90^\circ \) at the centre. Segment area \( = \) sector \( - \) triangle \( =\frac{\pi a^2}{4}-\frac{a^2}{2} \).
2. \( =\frac{a^2}2\left(\frac\pi2-1\right) \).

Teacher's Note:
1. Segment area = sector area minus triangle area.

Answer: (b) \( \frac13 \) sq unit

Explanation:
1. The curves meet at \( x=-1,0,1 \) and the region is symmetric about the y-axis.
2. Area \( =2\int_0^1(x-x^2)\,dx=2\left(\frac12-\frac13\right)=\frac13 \).

Teacher's Note:
1. Use the symmetry about the y-axis.

Answer: (a) \( \frac98 \) sq units

Explanation:
1. Solving \( x^2=x+2 \) gives \( x=-1 \) and \( x=2 \).
2. Area \( =\int_{-1}^2\left(\frac{x+2}{4}-\frac{x^2}{4}\right)dx=\frac14\cdot\frac92=\frac98 \).

Teacher's Note:
1. The line is above the parabola between the intersections.

Answer: (d) \( \frac{21}2 \) sq units

Explanation:
1. On \( [0,3] \) the parabola \( y=x^2+2 \) lies above the line \( y=x \).
2. Area \( =\int_0^3(x^2+2-x)\,dx=9+6-\frac92=\frac{21}2 \).

Teacher's Note:
1. Upper curve minus lower curve.

Answer: (a) 4 sq units

Explanation:
1. Area \( =\frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=\frac12|(-1)(1)+1(2)+3(-3)| \).
2. \( =\frac12|-1+2-9|=4 \) (integration of the three side lines gives the same value).

Teacher's Note:
1. Verify with the coordinate-geometry area formula.

Answer: (c) \( \frac\pi2 \) sq units

Explanation:
1. \( |x|\le|y| \) describes the two opposite angular sectors around the y-axis bounded by the lines \( y=\pm x \); together they cover half of the unit disc.
2. Area \( =\frac12\pi(1)^2=\frac\pi2 \).

Teacher's Note:
1. The lines \( y=\pm x \) split the disc into four equal sectors.

Answer: (d) \( -\frac34 \) sq unit

Explanation:
1. The intercepts give \( f(2)=0 \) and \( f(0)=1 \), and \( \int_0^2f(x)\,dx=\frac34 \).
2. Integrating by parts: \( \int_0^2xf'(x)\,dx=[xf(x)]_0^2-\int_0^2f(x)\,dx=0-\frac34=-\frac34 \).

Teacher's Note:
1. Use integration by parts with \( u=x \).