JEE Mathematics Area Under The Curve MCQs Set 03

Answer: (d) \( \dfrac{17\sqrt{17}}{6} \)

Explanation:
1. Intersection: \( x^2-1=3-x \), so \( x^2+x-4=0 \) with discriminant 17.
2. Area between a line and a parabola with \( |a|=1 \): \( \dfrac{(\sqrt D)^3}{6a^2}=\dfrac{17\sqrt{17}}{6} \).

Teacher's Note:
1. The area between a parabola \( y=ax^2+bx+c \) and a line meeting it at two points is \( \dfrac{D^{3/2}}{6a^2} \), with D the discriminant of the intersection equation.
2. Verify by direct integration if time permits.

Answer: (c) \( e^2-2 \)

Explanation:
1. \( e^x>|x-1| \) for all x, so area \( =\displaystyle\int_0^2e^x\,dx-\int_0^2|x-1|\,dx \).
2. \( \int_0^2e^xdx=e^2-1 \) and \( \int_0^2|x-1|dx=\tfrac12+\tfrac12=1 \).
3. Area \( =e^2-2 \).

Teacher's Note:
1. Split the integral of an absolute value at the point where the expression changes sign.
2. Check which curve lies above before subtracting.

Answer: (b) \( \dfrac43 \)

Explanation:
1. \( y=\pm|x|\sqrt{1-x^2} \), defined for \( |x|\le1 \). The curve is symmetric in both axes.
2. Area \( =4\displaystyle\int_0^1x\sqrt{1-x^2}\,dx \).
3. With \( u=1-x^2 \): \( \int_0^1x\sqrt{1-x^2}dx=\dfrac13 \). Area \( =\dfrac43 \).

Teacher's Note:
1. Use symmetry about both axes and compute one quarter.
2. A substitution \( u=1-x^2 \) handles the integral at once.

Answer: (c) \( 3-e \)

Explanation:
1. Intersection: \( \ln x=(\ln x)^2 \), so \( x=1 \) or \( x=e \). On \( (1,e) \), \( \ln x\ge(\ln x)^2 \).
2. \( \int_1^e\ln x\,dx=1 \) and \( \int_1^e(\ln x)^2dx=e-2 \).
3. Area \( =1-(e-2)=3-e \).

Teacher's Note:
1. On \( (1,e) \) we have \( 0\le\ln x\le1 \), so squaring makes the value smaller.
2. Use \( \int(\ln x)^2dx=x(\ln x)^2-2x\ln x+2x \).

Answer: (a) \( \dfrac{10}3 \)

Explanation:
1. At (1, 2): \( y'=\dfrac2y=1 \), so the normal has slope \( -1 \): \( x+y=3 \), meeting the x-axis at (3, 0).
2. Integrate with respect to y from 0 to 2: \( \displaystyle\int_0^2\left[(3-y)-\dfrac{y^2}{4}\right]dy \).
3. \( =6-2-\dfrac23=\dfrac{10}{3} \).

Teacher's Note:
1. Integrating along the y-axis is easier when the boundaries are given as x in terms of y.
2. The normal slope is the negative reciprocal of the tangent slope.

Answer: (c) 15

Explanation:
1. \( \int_0^4(f-g)=\int_0^2(f-g)+\int_2^4(f-g)=10 \).
2. \( \int_2^4(f-g)=-5 \).
3. So \( \int_0^2(f-g)=10+5=15 \), and since \( f>g \) there, the area is 15.

Teacher's Note:
1. Split the integral at the intersection point x = 2.
2. Area is positive, so use \( f-g \) where f is above g.

Answer: (a) \( \dfrac23 \)

Explanation:
1. A line \( y=mx \) touches the parabola when \( x^2-mx+1=0 \) has equal roots: \( m^2=4 \), so \( m=\pm2 \), touching at \( x=\pm1 \).
2. By symmetry area \( =2\displaystyle\int_0^1\left[(x^2+1)-2x\right]dx=2\int_0^1(x-1)^2dx=2\cdot\dfrac13 \).
3. Area \( =\dfrac23 \).

Teacher's Note:
1. The condition for tangency is a zero discriminant.
2. Symmetry about the y-axis halves the integral.

Answer: (d) \( \dfrac14 \)

Explanation:
1. The curves meet where \( x=\dfrac a2 \). \( S=\displaystyle\int_0^{a/2}\left(e^{a-x}-e^x\right)dx=e^a+1-2e^{a/2}=\left(e^{a/2}-1\right)^2 \).
2. As \( a\to0 \): \( e^{a/2}-1\sim\dfrac a2 \), so \( \dfrac S{a^2}\to\dfrac14 \).

Teacher's Note:
1. Recognise \( e^a-2e^{a/2}+1 \) as a perfect square.
2. Use \( e^h-1\sim h \) for small h.

Answer: (d) 64

Explanation:
1. \( f(x)=\tan^{-1}x \), \( f(1)=\dfrac\pi4 \), \( f'(1)=\dfrac12 \). The normal has slope \( -2 \): \( y-\dfrac\pi4=-2(x-1) \), meeting the x-axis at \( x=1+\dfrac\pi8 \).
2. Area \( =\displaystyle\int_0^1\tan^{-1}x\,dx+\dfrac12\cdot\dfrac\pi8\cdot\dfrac\pi4=\left(\dfrac\pi4-\ln\sqrt2\right)+\dfrac{\pi^2}{64} \).
3. Hence k = 64.

Teacher's Note:
1. The region splits into the area under the curve from 0 to 1 and a right triangle.
2. \( \int\tan^{-1}x\,dx \) is done by parts.

Answer: (c) \( \dfrac{11}4 \)

Explanation:
1. \( y=x^3-6x^2+11x-6 \) with antiderivative \( G=\dfrac{x^4}4-2x^3+\dfrac{11x^2}2-6x \).
2. \( G(0)=0,\ G(1)=-\dfrac94,\ G(2)=-2,\ G(3)=-\dfrac94 \).
3. The curve is negative on (0, 1), positive on (1, 2), negative on (2, 3): area \( =\dfrac94+\dfrac14+\dfrac14=\dfrac{11}4 \).

Teacher's Note:
1. Add the absolute values of the integrals over intervals where the curve changes sign.
2. Evaluate the antiderivative at the roots to save work.

Answer: (a) \( \dfrac{13}6 \)

Explanation:
1. Total area \( =\displaystyle\int_0^{3/2}(1+4x-x^2)dx=\dfrac32+\dfrac92-\dfrac98=\dfrac{39}8 \).
2. The area under the line from 0 to \( \dfrac32 \) is \( \dfrac12\cdot\dfrac32\cdot\dfrac{3m}2=\dfrac{9m}8 \) and it must equal half: \( \dfrac{39}{16} \).
3. \( m=\dfrac{13}6 \) (the line then ends at height 3.25, below the curve value 4.75).

Teacher's Note:
1. The line from the origin cuts off a triangle under it.
2. Check that the line stays under the curve on the interval.

Answer: (b) \( \dfrac{4-3\ln3}{2} \)

Explanation:
1. The first curve is \( y=x \) for \( x\le2 \) and \( y=4-x \) for \( x>2 \). It meets \( y=\dfrac3x \) at \( x=\sqrt3 \) and \( x=3 \).
2. Area \( =\displaystyle\int_{\sqrt3}^2\left(x-\dfrac3x\right)dx+\int_2^3\left(4-x-\dfrac3x\right)dx \).
3. \( =\left(\dfrac12-3\ln2+\dfrac32\ln3\right)+\left(\dfrac32-3\ln3+3\ln2\right)=2-\dfrac32\ln3=\dfrac{4-3\ln3}{2} \).

Teacher's Note:
1. Rewrite the modulus function piecewise before intersecting.
2. Add the two integrals and watch the cancellation of the \( \ln2 \) terms.

Answer: (d) \( xe^x+e^x \)

Explanation:
1. \( \displaystyle\int_0^{x_1}f(x)dx=x_1e^{x_1} \).
2. Differentiate with respect to \( x_1 \): \( f(x_1)=e^{x_1}+x_1e^{x_1} \).
3. So \( f(x)=xe^x+e^x \).

Teacher's Note:
1. Differentiate the area function to recover the curve.
2. Use the product rule on \( x_1e^{x_1} \).

Answer: (b) \( \dfrac{16}3 \)

Explanation:
1. With \( t=2x+1 \): \( 4x^2+4x+5=(t-1)^2+2(t-1)+5=t^2+4 \), so \( f(x)=x^2+4 \).
2. A line \( y=mx \) is tangent when \( x^2-mx+4=0 \) has equal roots: \( m=\pm4 \), at \( x=\pm2 \).
3. Area \( =2\displaystyle\int_0^2\left(x^2+4-4x\right)dx=2\int_0^2(x-2)^2dx=2\cdot\dfrac83=\dfrac{16}3 \).

Teacher's Note:
1. First recover f by expressing h in terms of g.
2. Then proceed as for any tangents from an external point.

Answer: (c) \( \dfrac12 \)

Explanation:
1. Area \( =\displaystyle\int_0^{1/c}(x^2-cx^3)dx=\dfrac1{3c^3}-\dfrac c4\cdot\dfrac1{c^4}=\dfrac1{12c^3} \).
2. \( \dfrac1{12c^3}=\dfrac23 \), so \( c^3=\dfrac18 \).
3. \( c=\dfrac12 \).

Teacher's Note:
1. On \( [0,1/c] \) the parabola lies above the cubic.
2. Solve the resulting equation for \( c^3 \) and take the real cube root.

Answer: (b) \( \dfrac{e^2-5e^{-2}}{4} \)

Explanation:
1. \( y\ge0 \) for \( x\le e \). \( \displaystyle\int_{1/e}^ex\,dx=\dfrac{e^2-e^{-2}}2 \).
2. \( \displaystyle\int x\ln x\,dx=\dfrac{x^2}2\ln x-\dfrac{x^2}4 \), so \( \int_{1/e}^ex\ln x\,dx=\dfrac{e^2}4+\dfrac34e^{-2} \).
3. Area \( =\dfrac{e^2-e^{-2}}2-\dfrac{e^2}4-\dfrac{3e^{-2}}4=\dfrac{e^2-5e^{-2}}4 \).

Teacher's Note:
1. Use integration by parts for \( x\ln x \).
2. Be careful with the lower limit \( 1/e \), where \( \ln x=-1 \).

Answer: (d) 8

Explanation:
1. \( x(x-3)^2=x \) gives \( x=0 \) or \( (x-3)^2=1 \), i.e. \( x=2,4 \).
2. On \( (0,2) \) the cubic is above the line; on \( (2,4) \) the line is above.
3. Area \( =\displaystyle\int_0^2\left[x(x-3)^2-x\right]dx+\int_2^4\left[x-x(x-3)^2\right]dx=4+4=8 \).

Teacher's Note:
1. Check which curve is on top in each interval between intersection points.
2. The two pieces happen to be equal here.

Answer: (c) \( e-1 \)

Explanation:
1. \( f'=ce^x \) with \( f'(0)=1 \) gives \( f'=e^x \), so \( f=e^x+k \) and \( f(0)=0 \) gives \( f=e^x-1 \).
2. The height above the line \( y=-1 \) is \( f+1=e^x \).
3. Area \( =\displaystyle\int_0^1e^xdx=e-1 \).

Teacher's Note:
1. Solve the differential conditions first to find f.
2. The area is measured from the line y = -1, so add 1 to f.

Answer: (d) 1

Explanation:
1. \( A(a)=\displaystyle\int_a^{2a}\left(\dfrac x6+\dfrac1{x^2}\right)dx=\dfrac{a^2}{4}+\dfrac1{2a} \).
2. \( A'(a)=\dfrac a2-\dfrac1{2a^2}=0 \) gives \( a^3=1 \), so \( a=1 \).
3. \( A''(1)>0 \), so this is a minimum.

Teacher's Note:
1. Express the area as a function of a and then use calculus to minimise it.
2. Confirm with the second derivative.

Answer: (d) 2

Explanation:
1. Both curves meet at \( y=\pm1 \) (x = 0). For \( |y|\le1 \), \( |y|\sqrt{1-y^2}\ge0\ge y^2-1 \).
2. Area \( =\displaystyle\int_{-1}^1\left[|y|\sqrt{1-y^2}-(y^2-1)\right]dy=2\int_0^1y\sqrt{1-y^2}\,dy+2\int_0^1(1-y^2)dy \).
3. \( =2\cdot\dfrac13+2\cdot\dfrac23=2 \).

Teacher's Note:
1. Integrate along the y-axis because both curves are given as x in terms of y.
2. Use symmetry in y to double the integral over \( [0,1] \).

Answer: (d) \( \dfrac{13}4 \)

Explanation:
1. \( f(1)=5 \), so \( g(5)=1 \) and \( g(0)=0 \). The area under g from 0 to 5 equals the area of the rectangle \( 5\times1 \) minus the area under f from 0 to 1.
2. \( \displaystyle\int_0^1(3x^3+2x)dx=\dfrac34+1=\dfrac74 \).
3. Area \( =5-\dfrac74=\dfrac{13}4 \).

Teacher's Note:
1. For an inverse function, \( \int_0^{b}g\,dy=b\,g(b)-\int_0^{g(b)}f\,dx \).
2. This avoids solving the cubic for the inverse.