JEE Mathematics Area Under The Curve MCQs Set 07

Answer: (b) \( \frac{19}{6} \)

Explanation:
1. The parabola \( y=(x-1)(x-4) \) and the line \( y=1-x \) meet where \( x^2-4x+3=0 \), i.e. at \( x=1 \) and \( x=3 \).
2. For \( 1\le x\le3 \) the region lies between \( y=1-x \) and \( y=0 \): \( \int_1^3(x-1)\,dx=2 \). For \( 3\le x\le4 \) it lies between the parabola and \( y=0 \): \( \int_3^4-(x^2-5x+4)\,dx=\frac76 \).
3. Total area \( =2+\frac76=\frac{19}6 \).

Teacher's Note:
1. Sketch the parabola, the line and the x-axis first; the lower boundary changes at x = 3.

Answer: (c) \( \frac43 \)

Explanation:
1. The curves are \( y=-2x^2 \) and \( y=1-3x^2 \). They meet where \( -2x^2=1-3x^2 \), i.e. \( x=\pm1 \).
2. Between these points \( y=1-3x^2 \) lies above \( y=-2x^2 \), so the area is \( \int_{-1}^{1}\left(1-x^2\right)dx \).
3. \( =2\left(1-\frac13\right)=\frac43 \).

Teacher's Note:
1. Find the intersection points first; the region is symmetric about the y-axis.

Answer: (c) 2

Explanation:
1. Integrate with respect to \(y\): \( x=e^y \) runs from \( x=0 \) (the y-axis) to the curve.
2. Area \( =\int_0^{\ln3}e^y\,dy=e^{\ln3}-1 \).
3. \( =3-1=2 \).

Teacher's Note:
1. When the region is bounded by the y-axis, integrating along y is easier.

Answer: (a) \( \frac{14}9 \)

Explanation:
1. In the first quadrant \( x=\frac{\sqrt y}{3} \).
2. Area \( =\int_1^4\frac{\sqrt y}3dy=\frac13\cdot\frac23\left[y^{3/2}\right]_1^4 \).
3. \( =\frac29(8-1)=\frac{14}9 \).

Teacher's Note:
1. Express x as a function of y and integrate between the two horizontal lines.

Answer: (c) \( 2\sqrt2-2 \)

Explanation:
1. \( \cos x-\sin x \) is positive on \( \left[0,\frac\pi4\right] \) and negative on \( \left[\frac\pi4,\frac\pi2\right] \).
2. \( \int_0^{\pi/4}(\cos x-\sin x)dx=\sqrt2-1 \) and \( \int_{\pi/4}^{\pi/2}(\sin x-\cos x)dx=\sqrt2-1 \).
3. The area is \( 2\sqrt2-2 \).

Teacher's Note:
1. Split the interval at the point where the expression inside the modulus changes sign.

Answer: (b) \( \frac{17}{27} \) square units

Explanation:
1. \( (1-x)^2=2x(1-x) \) at \( x=\frac13 \) and \( x^2=2x(1-x) \) at \( x=\frac23 \). So \( f=(1-x)^2 \) on \( \left[0,\frac13\right] \), \( f=2x(1-x) \) on \( \left[\frac13,\frac23\right] \), \( f=x^2 \) on \( \left[\frac23,1\right] \).
2. The three areas are \( \frac{19}{81},\ \frac{13}{81},\ \frac{19}{81} \).
3. Total \( =\frac{51}{81}=\frac{17}{27} \).

Teacher's Note:
1. Compare the three expressions pairwise to find where the maximum switches.

Answer: (a) \( \frac{(e+1)\pi}{1+\pi^2} \)

Explanation:
1. Area \( =\int_0^1e^y\sin\pi y\,dy \). Using \( \int e^{y}\sin\pi y\,dy=\frac{e^y(\sin\pi y-\pi\cos\pi y)}{1+\pi^2} \).
2. At \( y=1 \): \( \frac{e(0+\pi)}{1+\pi^2} \). At \( y=0 \): \( \frac{-\pi}{1+\pi^2} \).
3. Area \( =\frac{\pi e+\pi}{1+\pi^2}=\frac{(e+1)\pi}{1+\pi^2} \).

Teacher's Note:
1. Integrate with respect to y and use the standard formula for the integral of e^y sin(ky).

Answer: (c) \( \frac18 \)

Explanation:
1. For \( 0\le x<1 \), \( \{x\}=x \), so the curve is the line \( y=x \).
2. The region is the triangle with vertices \( (0,0),\ \left(\frac12,0\right),\ \left(\frac12,\frac12\right) \).
3. Area \( =\int_0^{1/2}x\,dx=\frac18 \).

Teacher's Note:
1. On [0, 1) the fractional part function is just y = x.

Answer: (a) \( \frac{81}2 \)

Explanation:
1. \( x=9y-y^3=y(3-y)(3+y) \) is zero at \( y=0,\pm3 \), so the curve meets the y-axis at those points.
2. The curve is odd in \(y\), so the two loops have equal area: \( 2\int_0^3\left(9y-y^3\right)dy \).
3. \( =2\left[\frac{9y^2}2-\frac{y^4}4\right]_0^3=2\left(\frac{81}2-\frac{81}4\right)=\frac{81}2 \).

Teacher's Note:
1. Use the symmetry of the odd function x(y) about the origin.

Answer: (a) \( \frac{4-3\ell n3}{2} \)

Explanation:
1. \( y=2-|2-x| \) is \( y=x \) for \( x\le2 \) and \( y=4-x \) for \( x\ge2 \). It meets \( y=\frac3x \) at \( x=\sqrt3 \) (left part) and \( x=3 \) (right part).
2. Area \( =\int_{\sqrt3}^2\left(x-\frac3x\right)dx+\int_2^3\left(4-x-\frac3x\right)dx=\left(\frac12-3\ln2+\frac32\ln3\right)+\left(\frac32-3\ln3+3\ln2\right) \).
3. \( =2-\frac32\ln3=\frac{4-3\ln3}2 \).

Teacher's Note:
1. Draw the tent-shaped graph and the hyperbola; the enclosed region lies between x = sqrt(3) and x = 3.

Answer: (c) II and IV only

Explanation:
1. The curve and the line meet at \( x=0 \) and \( x=1-m \).
2. The area between them is \( \left|\int_0^{1-m}\left(x-x^2-mx\right)dx\right|=\frac{|1-m|^3}{6} \).
3. Setting this equal to \( \frac92 \) gives \( |1-m|^3=27 \), so \( 1-m=\pm3 \), i.e. \( m=-2 \) or \( m=4 \).

Teacher's Note:
1. The area between a parabola and a chord is |discriminant-type length|^3 / 6.

Answer: (b) \( 6-4\ell n2 \)

Explanation:
1. For \( 0\le x\le\ln2 \) the region is bounded above by \( y=e^x \) (which is below 2) and below by \( y=0 \): area \( e^{\ln2}-1=1 \).
2. For \( \ln2\le x\le1 \) the strip runs from \( y=0 \) to \( y=2 \): area \( 2(1-\ln2) \). For \( 1\le x\le2 \) it runs from \( y=\ln x \) to 2: \( \int_1^2(2-\ln x)dx=3-2\ln2 \).
3. Total \( =1+2-2\ln2+3-2\ln2=6-4\ln2 \).

Teacher's Note:
1. Split the x-interval where the bounding curves change.

Answer: (b) 8

Explanation:
1. The curve consists of \( y=x^3 \) and \( y=-x^3 \), symmetric about the x-axis.
2. Area \( =2\int_0^2x^3dx \).
3. \( =2\cdot\frac{2^4}4=8 \).

Teacher's Note:
1. The two arms are mirror images in the x-axis, so double the area of one arm.

Answer: (d) \( \frac{3\pi a^2}8 \)

Explanation:
1. The curve is the astroid \( x^{2/3}+y^{2/3}=a^{2/3} \), symmetric in all four quadrants.
2. Area \( =4\int_0^{a}y\,dx=4\int_0^{\pi/2}a\sin^3t\cdot3a\cos^2t\sin t\,dt=12a^2\int_0^{\pi/2}\sin^4t\cos^2t\,dt \).
3. \( \int_0^{\pi/2}\sin^4t\cos^2t\,dt=\frac{\pi}{32} \), so the area is \( \frac{12\pi a^2}{32}=\frac{3\pi a^2}8 \).

Teacher's Note:
1. Use symmetry to take four times the first-quadrant area and a Beta/Wallis integral.

Answer: (b) 9

Explanation:
1. The line is \( y=\frac{x-3}2 \). It meets \( y=\sqrt x \) where \( (x-3)^2=4x \), i.e. \( x^2-10x+9=0 \), giving \( x=1 \) or \( 9 \). Only \( x=9 \) satisfies the line's equation (point \( (9,3) \)).
2. The region lies under \( y=\sqrt x \) from 0 to 9 and above the line from 3 to 9 (the line meets the x-axis at \( x=3 \)).
3. Area \( =\int_0^9\sqrt x\,dx-\int_3^9\frac{x-3}2dx=18-9=9 \).

Teacher's Note:
1. Check which intersection point actually lies on the line before using it as a limit.

Answer: (c) \( \frac{20\sqrt2}3 \)

Explanation:
1. For a given \(y\), the right-hand branches are \( x=\frac{\sqrt y}2 \) and \( x=3\sqrt y \), so the width between the parabolas on each side is \( \frac52\sqrt y \).
2. Using both sides: area \( =2\int_0^2\frac52\sqrt y\,dy=5\int_0^2\sqrt y\,dy \).
3. \( =5\cdot\frac23\cdot2\sqrt2=\frac{20\sqrt2}3 \).

Teacher's Note:
1. Integrate along y; the region is symmetric about the y-axis.

Answer: (d) \( \frac{16}3 \)

Explanation:
1. The curves meet at \( (0,0) \) and \( (4,4) \).
2. For \( 0\le x\le4 \), \( y=2\sqrt x \) lies above \( y=\frac{x^2}4 \).
3. Area \( =\int_0^4\left(2\sqrt x-\frac{x^2}4\right)dx=\frac{32}3-\frac{16}3=\frac{16}3 \).

Teacher's Note:
1. Use the standard result: the area between y^2 = 4ax and x^2 = 4ay is 16a^2/3.

Answer: (d) \( \frac32 \) square units

Explanation:
1. The lines \( y=x \) and \( y=\frac1x \) meet at \( (1,1) \).
2. For \( 0\le x\le1 \) the region lies under \( y=x \); for \( 1\le x\le e \) it lies under \( y=\frac1x \).
3. Area \( =\int_0^1x\,dx+\int_1^e\frac{dx}x=\frac12+1=\frac32 \).

Teacher's Note:
1. The upper boundary changes at the intersection point (1, 1).

Answer: (a) 9 square unit

Explanation:
1. Differentiating \( (y-2)^2=x-1 \): \( 2(y-2)\frac{dy}{dx}=1 \), so the slope at \( (2,3) \) is \( \frac12 \). The tangent is \( x=2y-4 \).
2. Integrate with respect to \(y\) from \( y=0 \) (the x-axis) to \( y=3 \): the parabola is \( x=1+(y-2)^2 \) on the right and the tangent on the left.
3. Area \( =\int_0^3\left[1+(y-2)^2-(2y-4)\right]dy=\int_0^3(y-3)^2dy=9 \).

Teacher's Note:
1. Write both boundaries as x in terms of y; the difference collapses to a perfect square.

Answer: (d) \( \frac43 \) square unit

Explanation:
1. The curves are \( x=-2y^2 \) and \( x=1-3y^2 \). They meet where \( y^2=1 \), i.e. \( y=\pm1 \).
2. Between these, \( x=1-3y^2 \) is to the right, so the width is \( 1-y^2 \).
3. Area \( =\int_{-1}^{1}\left(1-y^2\right)dy=\frac43 \).

Teacher's Note:
1. Both curves are parabolas opening along the x-axis, so integrate with respect to y.

Answer: (a) \( \frac\pi2+\frac43 \)

Explanation:
1. The parabola is \( x=1-y^2 \) and the circle gives \( x=\pm\sqrt{1-y^2} \). For \( |y|<1 \), \( 1-y^2\le\sqrt{1-y^2} \), so the region lies between \( x=-\sqrt{1-y^2} \) and \( x=1-y^2 \).
2. Area \( =\int_{-1}^{1}\left(1-y^2+\sqrt{1-y^2}\right)dy \).
3. \( \int_{-1}^1(1-y^2)dy=\frac43 \) and \( \int_{-1}^{1}\sqrt{1-y^2}dy=\frac\pi2 \), so the area is \( \frac\pi2+\frac43 \).

Teacher's Note:
1. Compare the parabola and the circle at a given y to see which boundary is on the right.