JEE Mathematics Application of Integrals MCQs Set 11

Answer: (c) \(\frac{16}{3}a^2\)

Explanation:
1. They meet at \((0,0)\) and \((4a,4a)\). Area \(=\int_0^{4a}\left(2\sqrt a\sqrt x-\frac{x^2}{4a}\right)dx\).
2. \(=2\sqrt a\cdot\frac23(4a)^{3/2}-\frac{(4a)^3}{12a}=\frac{32a^2}{3}-\frac{16a^2}{3}=\frac{16a^2}{3}\).

Teacher's Note:
1. Standard result: the area between \(y^2=4ax\) and \(x^2=4ay\) is \(\frac{16a^2}{3}\).
2. Key result: \(=2\sqrt a\cdot\frac23(4a)^{3/2}-\frac{(4a)^3}{12a}=\frac{32a^2}{3}-\frac{16a^2}{3}=\frac{16a^2}{3}\).

Answer: (d) \(\frac43(8+3\pi)\)

Explanation:
1. The circle \((x-4)^2+y^2=16\) meets the parabola at \((0,0)\) and \((4,4)\).
2. Area \(=\int_0^42\sqrt x\,dx+\int_4^8\sqrt{16-(x-4)^2}\,dx=\frac{32}{3}+4\pi=\frac43(8+3\pi)\).

Teacher's Note:
1. The second integral is a quarter of a circle of radius 4.
2. Key result: Area \(=\int_0^42\sqrt x\,dx+\int_4^8\sqrt{16-(x-4)^2}\,dx=\frac{32}{3}+4\pi=\frac43(8+3\pi)\).

Answer: (d) \(4\)

Explanation:
1. The curves meet at \(x=0\) and \(x=\frac{16}{m^2}\). Area \(=\int_0^{16/m^2}(4\sqrt x-mx)dx=\frac{128}{3m^3}\).
2. Setting \(\frac{128}{3m^3}=\frac23\) gives \(m^3=64\), so \(m=4\).

Teacher's Note:
1. Express the area through m and equate it to the given value.
2. Key result: Setting \(\frac{128}{3m^3}=\frac23\) gives \(m^3=64\), so \(m=4\).

Answer: (d) \(none of these\)

Explanation:
1. Area \(=\int_0^1x^2dx+\int_1^2(2-x)dx=\frac13+\frac12=\frac56\).
2. This is not among the three given values.

Teacher's Note:
1. Add the areas under the two pieces separately.
2. Key result: This is not among the three given values.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both integrals equal \(\frac{\pi a^2}{4}\). The first slices the quadrant vertically and the second horizontally.
2. R is true and it explains why the area is the same by either method; both statements hold.

Teacher's Note:
1. The circle is symmetric in x and y, so the two integrals are identical.
2. Key result: R is true and it explains why the area is the same by either method; both statements hold.

Answer: (c) \(9\)

Explanation:
1. The line \(y=\frac{x-3}{2}\) meets the x-axis at \(x=3\) and the curve at \(x=9\).
2. Area \(=\int_0^9\sqrt x\,dx-\int_3^9\frac{x-3}{2}dx=18-9=9\).

Teacher's Note:
1. Subtract the triangle under the line from the area under the curve.
2. Key result: Area \(=\int_0^9\sqrt x\,dx-\int_3^9\frac{x-3}{2}dx=18-9=9\).

Answer: (d) \(\frac{\pi^3-8}{4}\)

Explanation:
1. The quarter of the circle has area \(\frac{\pi\cdot\pi^2}{4}=\frac{\pi^3}{4}\).
2. The area under \(y=\sin x\) from \(0\) to \(\pi\) is \(2\).
3. Required area \(=\frac{\pi^3}{4}-2=\frac{\pi^3-8}{4}\).

Teacher's Note:
1. Subtract the area under the sine arch from the circular quadrant.
2. Key result: Required area \(=\frac{\pi^3}{4}-2=\frac{\pi^3-8}{4}\).

Answer: (a) \(8\)

Explanation:
1. The vertices are \((0,0)\), \((4,0)\) and \((4,-4)\).
2. Area \(=\frac12\cdot4\cdot4=8\).

Teacher's Note:
1. Find the three vertices from the pairs of lines.
2. Key result: Area \(=\frac12\cdot4\cdot4=8\).

Answer: (c) \(\frac{45}{4}\)

Explanation:
1. Integrate with respect to y: \(x=y^{1/3}\). Area \(=\int_1^8y^{1/3}dy=\frac34\left[y^{4/3}\right]_1^8=\frac34(16-1)=\frac{45}{4}\).

Teacher's Note:
1. When the strip is bounded by horizontal lines, integrate along the y-axis.
2. Key result: Integrate with respect to y: \(x=y^{1/3}\). Area \(=\int_1^8y^{1/3}dy=\frac34\left[y^{4/3}\right]_1^8=\frac34(16-1)=\frac{45}{4}\).

Answer: (c) \(7\)

Explanation:
1. Line AB: \(y=\frac52(x-2)\), line AC: \(y=\frac34(x-2)\) and line BC: \(y=9-x\).
2. Area \(=\int_2^4\left[\frac52-\frac34\right](x-2)dx+\int_4^6\left[(9-x)-\frac34(x-2)\right]dx=\frac74\cdot2+\left[\frac{21}{2}x-\frac78x^2\cdot2\right]_4^6=\frac72+\frac72=7\).
3. Check by the determinant formula: \(\frac12|2(5-3)+4(3-0)+6(0-5)|=7\).

Teacher's Note:
1. Split at the x-coordinate of the middle vertex and use upper minus lower line.
2. Key result: Check by the determinant formula: \(\frac12|2(5-3)+4(3-0)+6(0-5)|=7\).

Answer: (c) \(\frac54\)

Explanation:
1. Area \(=\int_{-1/2}^{1/2}(1+|x|)dx=1+2\int_0^{1/2}x\,dx=1+\frac14=\frac54\).

Teacher's Note:
1. The integrand is even, so double the integral over the positive side.
2. Key result: Area \(=\int_{-1/2}^{1/2}(1+|x|)dx=1+2\int_0^{1/2}x\,dx=1+\frac14=\frac54\).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The first-quadrant region under the cosine curve runs from \(x=0\) to \(x=\frac\pi2\), so A is true.
2. Its area is exactly the integral in R, so R explains A.

Teacher's Note:
1. The area under the curve equals its definite integral over the interval.
2. Key result: Its area is exactly the integral in R, so R explains A.

Answer: (b) \(\frac{ab(\pi-2)}{4}\)

Explanation:
1. The quadrant of the ellipse has area \(\frac{\pi ab}{4}\).
2. The triangle OAB has area \(\frac{ab}{2}\).
3. The segment between the arc and the chord is \(\frac{\pi ab}{4}-\frac{ab}{2}=\frac{ab(\pi-2)}{4}\).

Teacher's Note:
1. Subtract the triangle from the quadrant.
2. Key result: The segment between the arc and the chord is \(\frac{\pi ab}{4}-\frac{ab}{2}=\frac{ab(\pi-2)}{4}\).

Answer: (a) \(2\)

Explanation:
1. \(\int_0^\pi\sin x\,dx=\left[-\cos x\right]_0^\pi=1+1=2\).

Teacher's Note:
1. One arch of the sine curve has area 2.
2. Key result: \(\int_0^\pi\sin x\,dx=\left[-\cos x\right]_0^\pi=1+1=2\).

Answer: (c) \(\frac{8\sqrt8}{3}a^2\sqrt{\frac{b}{b-a}}\)

Explanation:
1. The parabolas meet at \(x_0=\frac{a(a+b)}{b-a}\), where \(y_0^2=4a(x_0+a)=\frac{8a^2b}{b-a}\).
2. Area \(=2\int_0^{y_0}\left[\left(\frac{y^2}{4b}+a\right)-\left(\frac{y^2}{4a}-a\right)\right]dy=4ay_0-\frac{(b-a)y_0^3}{6ab}\).
3. Using \(y_0^2=\frac{8a^2b}{b-a}\), the second term is \(\frac{4a}{3}y_0\), so the area is \(\frac{8a}{3}y_0=\frac{8\sqrt8}{3}a^2\sqrt{\frac{b}{b-a}}\).

Teacher's Note:
1. Write each parabola as x in terms of y and integrate the difference.
2. Key result: Using \(y_0^2=\frac{8a^2b}{b-a}\), the second term is \(\frac{4a}{3}y_0\), so the area is \(\frac{8a}{3}y_0=\frac{8\sqrt8}{3}a^2\sqrt{\frac{b}{b-a}}\).

Answer: (a) \(\frac{4\sqrt2}{3}\)

Explanation:
1. A tangent \(y=m(x-1)\) gives \(x^2-(2+m)x+3+m=0\) with zero discriminant: \(m^2=8\), so \(m=\pm2\sqrt2\). The points of contact are \(x=1\pm\sqrt2\).
2. The area between a parabola \(y=px^2+\dots\) and two tangents is \(\frac{|p|(x_2-x_1)^3}{12}\). Here \(\frac{(2\sqrt2)^3}{12}=\frac{16\sqrt2}{12}=\frac{4\sqrt2}{3}\).

Teacher's Note:
1. The parabola, with two tangents from an external point, encloses \(\frac{|p|\,d^3}{12}\) where d is the distance between the points of contact.
2. Key result: The area between a parabola \(y=px^2+\dots\) and two tangents is \(\frac{|p|(x_2-x_1)^3}{12}\). Here \(\frac{(2\sqrt2)^3}{12}=\frac{16\sqrt2}{12}=\frac{4\sqrt2}{3}\).

Answer: (d) \(3\)

Explanation:
1. \(\cos x\ge0\) on \([0,\frac\pi2]\) and \(\le0\) on \([\frac\pi2,\frac{3\pi}{2}]\).
2. Area \(=\int_0^{\pi/2}\cos x\,dx+\left|\int_{\pi/2}^{3\pi/2}\cos x\,dx\right|=1+2=3\).

Teacher's Note:
1. Take the absolute value of each piece where the function changes sign.
2. Key result: Area \(=\int_0^{\pi/2}\cos x\,dx+\left|\int_{\pi/2}^{3\pi/2}\cos x\,dx\right|=1+2=3\).

Answer: (b) \(12\pi\)

Explanation:
1. The area of an ellipse is \(\pi ab\) with \(a=4\), \(b=3\).
2. Area \(=12\pi\).

Teacher's Note:
1. Area of the ellipse is \(\pi ab\), where a and b are the semi-axes.
2. Key result: Area \(=12\pi\).

Answer: (d) \(none of these\)

Explanation:
1. The lines meet at \(\left(\frac23,-\frac23\right)\). The line \(y=2x-2\) meets the x-axis at \((1,0)\) and \(y=-x\) passes through the origin.
2. The region is the triangle with vertices \((0,0)\), \((1,0)\) and \(\left(\frac23,-\frac23\right)\).
3. Area \(=\frac12\cdot1\cdot\frac23=\frac13\), which is not among the first three options.

Teacher's Note:
1. Find the three vertices of the triangle and use half base times height.
2. Key result: Area \(=\frac12\cdot1\cdot\frac23=\frac13\), which is not among the first three options.

Answer: (c) A is true, but R is false.

Explanation:
1. Take \(x=1+\frac{y^2}{4a^2}\). The area between the curve and the line \(x=1\) is \(\int_0^{4a}\frac{y^2}{4a^2}dy=\frac{64a^3}{12a^2}=\frac{16a}{3}\), so A is true.
2. The curve \(y=x^2-x+2\) and the line meet where \(x^2-2x=0\), i.e. \(x=0,2\). The area is \(\int_0^2(2x-x^2)dx=\frac43\), not \(\frac83\), so R is false.

Teacher's Note:
1. Check each statement by direct integration, not by memory.
2. Key result: The curve \(y=x^2-x+2\) and the line meet where \(x^2-2x=0\), i.e. \(x=0,2\). The area is \(\int_0^2(2x-x^2)dx=\frac43\), not \(\frac83\), so R is false.

Answer: (c) \(\frac{2\pi}{3}-\frac{\sqrt3}{2}\)

Explanation:
1. The circles meet at \(x=\frac12\), \(y=\pm\frac{\sqrt3}{2}\). By symmetry the common area is \(4\int_{1/2}^1\sqrt{1-x^2}\,dx\).
2. \(\int\sqrt{1-x^2}dx=\frac12\left[x\sqrt{1-x^2}+\sin^{-1}x\right]\), so \(\int_{1/2}^1=\frac12\left[\frac\pi2-\frac{\sqrt3}{4}-\frac\pi6\right]=\frac12\left[\frac\pi3-\frac{\sqrt3}{4}\right]\).
3. Multiplying by 4 gives \(\frac{2\pi}{3}-\frac{\sqrt3}{2}\).

Teacher's Note:
1. The overlap of two unit circles whose centres are 1 apart is a lens.
2. Key result: Multiplying by 4 gives \(\frac{2\pi}{3}-\frac{\sqrt3}{2}\).

Answer: (d) \(4\)

Explanation:
1. The lines meet at \((-1,-1)\) and \((2,0)\); the vertices of the region are \((-1,-1)\), \((0,-2)\), \((2,0)\) and \((1,1)\).
2. By the shoelace formula, the area is \(\frac12|2+4+2+0|=4\).

Teacher's Note:
1. Locate all corner points of the region, then use the shoelace formula or split the area.
2. Key result: By the shoelace formula, the area is \(\frac12|2+4+2+0|=4\).

Answer: (a) \(2\sqrt3\)

Explanation:
1. The tangent is \(x+\sqrt3y=4\), meeting the x-axis at \((4,0)\). The normal passes through the origin.
2. The triangle has vertices \((0,0)\), \((4,0)\) and \((1,\sqrt3)\). Area \(=\frac12\cdot4\cdot\sqrt3=2\sqrt3\).

Teacher's Note:
1. The normal to a circle always passes through its centre.
2. Key result: The triangle has vertices \((0,0)\), \((4,0)\) and \((1,\sqrt3)\). Area \(=\frac12\cdot4\cdot\sqrt3=2\sqrt3\).

Answer: (d) \(\frac32\)

Explanation:
1. The line \(y=x\) meets \(y=\frac1x\) at \((1,1)\).
2. Area \(=\int_0^1x\,dx+\int_1^e\frac{dx}{x}=\frac12+1=\frac32\).

Teacher's Note:
1. Split the region at the intersection \(x=1\).
2. Key result: Area \(=\int_0^1x\,dx+\int_1^e\frac{dx}{x}=\frac12+1=\frac32\).

Answer: (c) \(\frac13\)

Explanation:
1. They meet at \(x=-1,0,1\). By symmetry the area is \(2\int_0^1(x-x^2)dx\).
2. \(=2\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=2\cdot\frac16=\frac13\).

Teacher's Note:
1. Use the symmetry about the y-axis to integrate on \([0,1]\) only.
2. Key result: \(=2\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=2\cdot\frac16=\frac13\).

Answer: (c) \(4\)

Explanation:
1. For \(y\ge0\), \(y=6\sqrt x\). Area \(=\int_0^16\sqrt x\,dx=6\cdot\frac23=4\).

Teacher's Note:
1. Use only the branch above the x-axis.
2. Key result: For \(y\ge0\), \(y=6\sqrt x\). Area \(=\int_0^16\sqrt x\,dx=6\cdot\frac23=4\).

Answer: (c) \(9\)

Explanation:
1. \(\int_0^3x^2dx=\frac{27}{3}=9\).

Teacher's Note:
1. The curve touches the x-axis at the origin, which gives the lower limit.
2. Key result: \(\int_0^3x^2dx=\frac{27}{3}=9\).

Answer: (d) A is false, but R is true.

Explanation:
1. Area \(=\int_0^\pi2\sin x\,dx=4\), so A is false.
2. The two curves are mirror images in the x-axis, so R is true.

Teacher's Note:
1. Double the area of one curve above the x-axis when the region is symmetric.
2. Key result: The two curves are mirror images in the x-axis, so R is true.

Answer: (c) \(6\)

Explanation:
1. For \(y\in[-1,1]\), \(x=2y+3>0\). Area \(=\int_{-1}^1(2y+3)dy=0+6=6\).

Teacher's Note:
1. Integrate x with respect to y between the horizontal lines.
2. Key result: For \(y\in[-1,1]\), \(x=2y+3>0\). Area \(=\int_{-1}^1(2y+3)dy=0+6=6\).

Answer: (b) \(\frac{21}{2}\)

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

Teacher's Note:
1. Upper curve minus lower curve, integrated between the given vertical lines.
2. Key result: \(=9+6-\frac92=\frac{21}{2}\).