JEE Mathematics Application of Integrals MCQs Set 09

Answer: (c) \( \dfrac{59}{6} \)

Explanation:
1. The curve \( y=x^2+3x \) meets \( y=4 \) at \( x=1 \) (for \( x\ge0 \)). For \( 0\le x\le1 \) the upper limit of \( y \) is the parabola, and for \( 1\le x\le3 \) it is \( y=4 \).
2. Area \( =\int_0^1(x^2+3x)\,dx+\int_1^3 4\,dx=\frac13+\frac32+8=\frac{59}{6} \). Answer: (c)

Teacher's Note:
1. Find where the parabola crosses the line \( y=4 \).

Answer: (c) \( 4\left(\dfrac4{25}\right)^{1/3} \)

Explanation:
1. \( A(\alpha)=2\int_0^\alpha\sqrt x\,dx=\dfrac43\alpha^{3/2} \).
2. \( \dfrac{\lambda^{3/2}}{4^{3/2}}=\dfrac25 \), so \( \left(\dfrac\lambda4\right)^3=\dfrac4{25} \) and \( \lambda=4\left(\dfrac4{25}\right)^{1/3} \). Answer: (c)

Teacher's Note:
1. Find the area as a function of \( \alpha \).

Answer: (b) \( \dfrac92 \)

Explanation:
1. The curves meet where \( x^2=x+2 \), that is \( x=-1,2 \).
2. Area \( =\int_{-1}^2(x+2-x^2)\,dx=\dfrac92 \). Answer: (b)

Teacher's Note:
1. Find the intersection points first.

Answer: (d) \( 18 \)

Explanation:
1. The parabola \( y^2=2x \) meets \( x=y+4 \) where \( y^2-2y-8=0 \), so \( y=-2,4 \).
2. Area \( =\int_{-2}^4\left(y+4-\dfrac{y^2}{2}\right)dy=\left[\dfrac{y^2}2+4y-\dfrac{y^3}6\right]_{-2}^{4}=18 \). Answer: (d)

Teacher's Note:
1. Integrate with respect to \( y \).

Answer: (c) square of side length \( 2\sqrt2 \) units

Explanation:
1. The four lines \( x-y=\pm2 \), \( x+y=\pm2 \) meet at \( (\pm2,0) \) and \( (0,\pm2) \).
2. These are the vertices of a square (perpendicular diagonals of equal length 4), with side \( \sqrt{2^2+2^2}=2\sqrt2 \) and area 8. Answer: (c)

Teacher's Note:
1. Find the four vertices.

Answer: (d) \( \dfrac32-\dfrac1{\log_e2} \)

Explanation:
1. In the first quadrant \( y=x+1 \) meets \( y=2^x \) at \( x=0 \) and \( x=1 \), and the line is above the curve between them.
2. Area \( =\int_0^1(x+1-2^x)\,dx=\dfrac12+1-\dfrac1{\log_e2}=\dfrac32-\dfrac1{\log_e2} \). Answer: (d)

Teacher's Note:
1. Compare the line and the exponential on \( [0,1] \).

Answer: (b) \( 6 \)

Explanation:
1. The parabola meets \( x+y=1 \) where \( x^2-6x+1=0 \), so \( x=3-2\sqrt2 \).
2. Area \( =\int_0^{3-2\sqrt2}2\sqrt x\,dx+\int_{3-2\sqrt2}^1(1-x)\,dx=\dfrac{8\sqrt2}3-\dfrac{10}3 \), giving \( a=\dfrac83 \), \( b=-\dfrac{10}3 \) and \( a-b=6 \). Answer: (b)

Teacher's Note:
1. Find the intersection with \( x+y=1 \) and split the area.

Answer: (c) \( 24 \)

Explanation:
1. They meet at \( x=0 \) and \( x=\dfrac4\lambda \).
2. Area \( =\int_0^{4/\lambda}(2\sqrt{\lambda x}-\lambda x)dx=\dfrac{32}{3\lambda}-\dfrac8\lambda=\dfrac{8}{3\lambda} \), and \( \dfrac8{3\lambda}=\dfrac19 \) gives \( \lambda=24 \). Answer: (c)

Teacher's Note:
1. Find the intersection points, then integrate.

Answer: (a) \( \dfrac83 \)

Explanation:
1. The tangent at \( (2,3) \) is \( y=4x-5 \).
2. Area \( =\int_0^2[(x^2-1)-(4x-5)]\,dx=\int_0^2(x-2)^2dx=\dfrac83 \). Answer: (a)

Teacher's Note:
1. Write the tangent first.

Answer: (a) \( 2 \)

Explanation:
1. For \( x\ge0 \) the bound is \( x^2+1 \), and for \( x\lt0 \) it is \( 1-x^2 \); both are non-negative on \( [-1,1] \).
2. Area \( =\int_{-1}^0(1-x^2)dx+\int_0^1(1+x^2)dx=\dfrac23+\dfrac43=2 \). Answer: (a)

Teacher's Note:
1. Split at \( x=0 \).

Answer: (b) \( \dfrac1{\sqrt3} \)

Explanation:
1. The curves meet at \( (0,0) \) and \( \left(\dfrac1k,\dfrac1k\right) \).
2. Area \( =\int_0^{1/k}\left(\sqrt{x/k}-kx^2\right)dx=\dfrac{2}{3k^2}-\dfrac1{3k^2}=\dfrac1{3k^2} \). Setting it equal to 1 gives \( k=\dfrac1{\sqrt3} \). Answer: (b)

Teacher's Note:
1. Find the second intersection point.

Answer: (b) \( \dfrac98 \)

Explanation:
1. Substituting \( y=\frac{x+2}4 \): \( x^2-x-2=0 \), so \( x=-1,2 \).
2. Area \( =\int_{-1}^2\left(\dfrac{x+2}4-\dfrac{x^2}4\right)dx=\dfrac14\cdot\dfrac92=\dfrac98 \). Answer: (b)

Teacher's Note:
1. Find the points of intersection.

Answer: (d) \( \dfrac{37}{24} \)

Explanation:
1. The tangent at \( (2,5) \) is \( y=4x-3 \), meeting the \( x\text{-axis} \) at \( \left(\frac34,0\right) \).
2. Area \( =\int_0^2(x^2+1)\,dx-\dfrac12\left(2-\dfrac34\right)5=\dfrac{14}3-\dfrac{25}8=\dfrac{37}{24} \). Answer: (d)

Teacher's Note:
1. Subtract the triangle under the tangent from the area under the parabola.

Answer: (a) \( \dfrac{15}2 \)

Explanation:
1. On \( [0,3] \) the parabola is above the line, since \( x^2-x+1\gt0 \).
2. Area \( =\int_0^3(x^2-x+1)\,dx=9-\dfrac92+3=\dfrac{15}2 \). Answer: (a)

Teacher's Note:
1. Subtract the lower curve from the upper curve.

Answer: (a) \( \dfrac12(\sqrt3-1) \)

Explanation:
1. The roots are \( \alpha=\dfrac\pi6 \) and \( \beta=\dfrac\pi3 \). Also \( (g\circ f)(x)=\cos\left((\sqrt x)^2\right)=\cos x \), which is positive on \( [\pi/6,\pi/3] \).
2. Area \( =\int_{\pi/6}^{\pi/3}\cos x\,dx=\dfrac{\sqrt3}2-\dfrac12=\dfrac12(\sqrt3-1) \). Answer: (a)

Teacher's Note:
1. Find the roots and simplify \( g\circ f \).

Answer: (d) \( \dfrac52 \)

Explanation:
1. The line \( x+y=3 \) meets \( y=1+\sqrt x \) at \( (1,2) \) and the parabola \( x^2=4y \) at \( (2,1) \).
2. Area \( =\int_0^1\left(1+\sqrt x-\dfrac{x^2}4\right)dx+\int_1^2\left(3-x-\dfrac{x^2}4\right)dx=\dfrac{19}{12}+\dfrac{11}{12}=\dfrac52 \). Answer: (d)

Teacher's Note:
1. Find the three boundary intersection points and split at \( x=1 \).

Answer: (b) \( \pi-\dfrac83 \)

Explanation:
1. The circle \( (x-2)^2+y^2=4 \) and the parabola \( y^2=2x \) meet at \( (0,0) \) and \( (2,2) \).
2. The region is the quarter circle (area \( \pi \)) minus the area under the parabola: \( \pi-\int_0^2\sqrt{2x}\,dx=\pi-\dfrac83 \). Answer: (b)

Teacher's Note:
1. Sketch the circle and the parabola.

Answer: (d) \( \dfrac9{32} \)

Explanation:
1. The line \( y=4x-1 \) meets \( y^2=2x \) where \( 16x^2-10x+1=0 \), so \( x=\frac12,\frac18 \), giving the points \( \left(\frac12,1\right) \) and \( \left(\frac18,-\frac12\right) \).
2. Integrating with respect to \( y \): Area \( =\int_{-1/2}^1\left(\dfrac{y+1}4-\dfrac{y^2}2\right)dy=\dfrac9{32} \). Answer: (d)

Teacher's Note:
1. Use horizontal strips between \( x=y^2/2 \) and \( x=(y+1)/4 \).

Answer: (a) \( \dfrac\pi2+\dfrac43 \)

Explanation:
1. For \( x\le0 \) the circle is the tighter bound (half disc, area \( \dfrac\pi2 \)); for \( 0\le x\le1 \) the parabola \( y^2\le1-x \) lies inside the circle.
2. Area \( =\dfrac\pi2+2\int_0^1(1-y^2)\,dy=\dfrac\pi2+\dfrac43 \). Answer: (a)

Teacher's Note:
1. Compare the circle and the parabola on each side of the y-axis.

Answer: (a) \( 9 \)

Explanation:
1. The curves meet where \( 2\sqrt x-x+3=0 \), giving \( \sqrt x=3 \), that is \( (9,3) \).
2. Using strips in \( y \): Area \( =\int_0^3\left[(2y+3)-y^2\right]dy=9+9-9=9 \). Answer: (a)

Teacher's Note:
1. Find the intersection and integrate with respect to \( y \).

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

Explanation:
1. By symmetry take twice the first-quadrant area: the parabolas are \( x=\dfrac{\sqrt y}2 \) and \( x=3\sqrt y \).
2. Area \( =2\int_0^2\left(3\sqrt y-\dfrac{\sqrt y}2\right)dy=5\cdot\dfrac23\cdot2^{3/2}=\dfrac{20\sqrt2}3 \). Answer: (c)

Teacher's Note:
1. Integrate the difference of the two \( x \)-values from \( y=0 \) to \( y=2 \).

Answer: (b) \( \dfrac32 \) sq units

Explanation:
1. \( y=x \) meets \( y=1/x \) at \( (1,1) \). For \( 0\le x\le1 \) the region is the triangle under \( y=x \) (area \( \dfrac12 \)).
2. For \( 1\le x\le e \) the region lies under \( y=\dfrac1x \): \( \int_1^e\dfrac{dx}x=1 \). Total area \( =\dfrac32 \). Answer: (b)

Teacher's Note:
1. Split at the point where the line meets the hyperbola.

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

Explanation:
1. The curves meet at \( (0,0) \) and \( (4,4) \).
2. Area \( =\int_0^4\left(2\sqrt x-\dfrac{x^2}4\right)dx=\dfrac{32}3-\dfrac{16}3=\dfrac{16}3 \). Answer: (c)

Teacher's Note:
1. Find the intersection points, then subtract the lower curve.

Answer: (a) \( (4\sqrt2-2) \) sq units

Explanation:
1. The curves cross at \( x=\dfrac\pi4 \) and \( x=\dfrac{5\pi}4 \), so split the area there.
2. Area \( =\int_0^{\pi/4}(\cos x-\sin x)dx+\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)dx+\int_{5\pi/4}^{3\pi/2}(\cos x-\sin x)dx=(\sqrt2-1)+2\sqrt2+(\sqrt2-1)=4\sqrt2-2 \). Answer: (a)

Teacher's Note:
1. Find the crossing points in the interval.

Answer: (b) \( 9 \) sq units

Explanation:
1. Differentiating, \( 2(y-2)\,dy=dx \), so at \( (2,3) \), \( \dfrac{dx}{dy}=2 \) and the tangent is \( x=2y-4 \), meeting the \( x\text{-axis} \) at \( (-4,0) \).
2. Area \( =\int_0^3\left[(y-2)^2+1-(2y-4)\right]dy=\int_0^3(y-3)^2dy=9 \). Answer: (b)

Teacher's Note:
1. Find the tangent and integrate with respect to \( y \).

Answer: (a) \( \dfrac43 \) sq units

Explanation:
1. They meet at \( y=\pm1 \), \( x=-2 \).
2. Area \( =\int_{-1}^1\left[(1-3y^2)-(-2y^2)\right]dy=\int_{-1}^1(1-y^2)\,dy=\dfrac43 \). Answer: (a)

Teacher's Note:
1. Find the intersection points and integrate in \( y \).

Answer: (c) \( \dfrac16 \) sq unit

Explanation:
1. The curves meet at \( (0,0) \) and \( (1,1) \); for \( x\ge0 \) and \( y\ge0 \) the parabola branch \( y=\sqrt x \) lies above \( y=x \). For \( x\lt0 \) the parabola has no points.
2. Area \( =\int_0^1(\sqrt x-x)\,dx=\dfrac23-\dfrac12=\dfrac16 \). Answer: (c)

Teacher's Note:
1. Check which side of the y-axis the curves bound a region.

Answer: (d) \( 1 \) sq unit

Explanation:
1. The curve meets the \( x\text{-axis} \) at \( x=1-e \) and the \( y\text{-axis} \) at \( y=1 \); the region lies between \( x=1-e \) and \( x=0 \).
2. Area \( =\int_{1-e}^0\log_e(x+e)\,dx=\int_1^e\ln t\,dt=[t\ln t-t]_1^e=1 \). Answer: (d)

Teacher's Note:
1. Find the intercepts and substitute \( t=x+e \).

Answer: (a) \( 1:1:1 \)

Explanation:
1. \( S_3=\int_0^4\dfrac{x^2}4dx=\dfrac{16}3 \), and by symmetry \( S_1=S_3=\dfrac{16}3 \).
2. \( S_2+S_3=\int_0^4 2\sqrt x\,dx=\dfrac{32}3 \), so \( S_2=\dfrac{16}3 \). The ratio is \( 1:1:1 \). Answer: (a)

Teacher's Note:
1. Use the symmetry of the two parabolas about \( y=x \).

Answer: (a) \( 1 \) sq unit

Explanation:
1. The graph is a V with vertex at \( (2,0) \), so on \( [1,2] \) it is \( 2-x \) and on \( [2,3] \) it is \( x-2 \).
2. Area \( =\int_1^2(2-x)dx+\int_2^3(x-2)dx=\dfrac12+\dfrac12=1 \). Answer: (a)

Teacher's Note:
1. Split the absolute value at \( x=2 \).

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

Explanation:
1. The graphs meet at \( (-1,2) \) and \( (2,1) \), and with the vertices \( (1,0) \) and \( (0,3) \) they form a rectangle with sides \( \sqrt2 \) and \( 2\sqrt2 \).
2. Area \( =\sqrt2\cdot2\sqrt2=4 \). Answer: (c)

Teacher's Note:
1. Find the four vertices of the region.

Answer: (a) \( \dfrac92 \) sq units

Explanation:
1. The curves meet where \( 2x-x^2=-x \), that is \( x=0 \) and \( x=3 \).
2. Area \( =\int_0^3(3x-x^2)\,dx=\dfrac{27}2-9=\dfrac92 \). Answer: (a)

Teacher's Note:
1. Find the intersection points first.