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Area Under Curves

JEE Mathematics · 107 questions · Page 9 of 11 · Click an option or "Show Solution" to reveal answer

Q81
The area of the region {(x,y):y2≤4x,x0,x≠3}\left\{(x, y): y^2 \leq 4 x, x0, x \neq 3\right\} is
A 323\dfrac{32}{3}
B 163\dfrac{16}{3}
C 83\dfrac{8}{3}
D 643\dfrac{64}{3}
Correct Answer
Option A
Solution
y2≤4x,x<4xy(x−1)(x−2)(x−3)(x−4)>0 Case - I:y>0x(x−1)(x−2)(x−3)(x−4)>0x∈(0,1)∪(2,3) Case − II : y<0 x(x−1)(x−2)(x−3)(x−4)<0,x∈(1,2)∪(3,4)\begin{aligned} & y^2 \leq 4 x, x<4 \\ & \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0 \\ & \text{ Case - I}: y>0 \\ & \frac{x(x-1)(x-2)}{(x-3)(x-4)}>0 \\ & x \in(0,1) \cup(2,3) \\ & \text{ Case }- \text{ II : y<0 } \\ & \frac{x(x-1)(x-2)}{(x-3)(x-4)}<0, x \in(1,2) \cup(3,4) \end{aligned}
 Area =2∫04xdx=2⋅23[x3/2]04=323\begin{aligned} & \text{ Area }=2 \int\limits_0^4 \sqrt{x} d x \\ & =2 \cdot \frac{2}{3}\left[x^{3 / 2}\right]_0^4=\frac{32}{3} \end{aligned}
Q82
The area (in square units) of the region bounded by the parabola y2=4(x−2)y^2=4(x-2) and the line y=2x−8y=2 x-8, is :
A 7
B 8
C 9
D 6
Correct Answer
Option C
Solution

Area along yy-axis

∫−24[(y+82)−(y2+84)]dy=y24+4y−y312−2y∣−24\begin{aligned} & \int\limits_{-2}^4\left[\left(\frac{y+8}{2}\right)-\left(\frac{y^2+8}{4}\right)\right] d y \\\\ & =\frac{y^2}{4}+4 y-\frac{y^3}{12}-\left.2 y\right|_{-2} ^4 \end{aligned}

=(4+16−163−8)−(1−8+812+4)=(12−163)−(812−3)=15−(64+812)=15−6=9 sq. unit \begin{aligned} & =\left(4+16-\dfrac{16}{3}-8\right)-\left(1-8+\dfrac{8}{12}+4\right) \\\\ & =\left(12-\dfrac{16}{3}\right)-\left(\dfrac{8}{12}-3\right)=15-\left(\dfrac{64+8}{12}\right) \\\\ & =15-6=9 \text{ sq. unit }\end{aligned}

Q83
The area (in square units) of the region enclosed by the ellipse x2+3y2=18x^2+3 y^2=18 in the first quadrant below the line y=xy=x is
A 3π+1\sqrt{3} \pi+1
B 3π\sqrt{3} \pi
C 3π−34\sqrt{3} \pi-\dfrac{3}{4}
D 3π+34\sqrt{3} \pi+\dfrac{3}{4}
Correct Answer
Option B
Solution
 Area =∫032xdx+∫3/23218−x23dx=12(x2)032+13[x218−x2+9sin⁡−1(x32)]3232=12(92)+13[9sin⁡−1(1)−322⋅332−9sin⁡−1(12)]=94+13(9π2−934−9π6)=3π\begin{aligned} & \text{ Area }=\int\limits_0^{3 \sqrt{2}} x d x+\int\limits_{3 / \sqrt{2}}^{3 \sqrt{2}} \sqrt{\frac{18-x^2}{3}} d x \\ & =\frac{1}{2}\left(x^2\right)_0^{3 \sqrt{2}}+\frac{1}{\sqrt{3}}\left[\frac{x}{2} \sqrt{18-x^2}+9 \sin ^{-1}\left(\frac{x}{3 \sqrt{2}}\right)\right]_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \\ & =\frac{1}{2}\left(\frac{9}{2}\right)+\frac{1}{\sqrt{3}}\left[9 \sin ^{-1}(1)-\frac{3}{2 \sqrt{2}} \cdot \frac{3 \sqrt{3}}{\sqrt{2}}-9 \sin ^{-1}\left(\frac{1}{2}\right)\right] \\ & =\frac{9}{4}+\frac{1}{\sqrt{3}}\left(\frac{9 \pi}{2}-\frac{9 \sqrt{3}}{4}-\frac{9 \pi}{6}\right)=\sqrt{3} \pi \end{aligned}
Q84
The area of the region, inside the circle (x−23)2+y2=12(x-2 \sqrt{3})^2+y^2=12 and outside the parabola y2=23xy^2=2 \sqrt{3} x is :
A 3π−83 \pi-8
B 6π−86 \pi-8
C 3π+83 \pi+8
D 6π−166 \pi-16
Correct Answer
Option D
Solution

 Required area =2∫023(43x−x2−23x)dx=2∫023(12−(x−23)2−23x)dx\begin{aligned} & \text{ Required area }=2 \int_0^{2 \sqrt{3}}\left(\sqrt{4 \sqrt{3} x-x^2}-\sqrt{2 \sqrt{3} x}\right) d x \\\\ & =2 \int_0^{2 \sqrt{3}}\left(\sqrt{12-(x-2 \sqrt{3})^2}-\sqrt{2 \sqrt{3} x}\right) d x\end{aligned} =2[x−23212−(x−23)2+122sin⁡−1(x−2323)−23x3232]023\begin{array}{r}=2\left[\dfrac{x-2 \sqrt{3}}{2} \sqrt{12-(x-2 \sqrt{3})^2}+\dfrac{12}{2} \sin ^{-1}\left(\dfrac{x-2 \sqrt{3}}{2 \sqrt{3}}\right)\right. \left.-\dfrac{\sqrt{2 \sqrt{3}} x^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right]_0^{2 \sqrt{3}}\end{array} =2{3π−8}=6π−16 sq. units. \begin{aligned} & =2\{3 \pi-8\} \\\\ & =6 \pi-16 \text{ sq. units. }\end{aligned}

Q85
The parabola y2=4xy^2=4 x divides the area of the circle x2+y2=5x^2+y^2=5 in two parts. The area of the smaller part is equal to :
A 23+5sin⁡−1(25)\dfrac{2}{3}+5 \sin ^{-1}\left(\dfrac{2}{\sqrt{5}}\right)
B 23+5sin⁡−1(25)\dfrac{2}{3}+\sqrt{5} \sin ^{-1}\left(\dfrac{2}{\sqrt{5}}\right)
C 13+5sin⁡−1(25)\dfrac{1}{3}+5 \sin ^{-1}\left(\dfrac{2}{\sqrt{5}}\right)
D 13+5sin⁡−1(25)\dfrac{1}{3}+\sqrt{5} \sin ^{-1}\left(\dfrac{2}{\sqrt{5}}\right)
Correct Answer
Option A
Solution

The point of intersection of

y2=4xy^2=4 x

and

x2+y2=5x^2+y^2=5

are

(1,2)(1,2)

and

(1,−2)(1,-2)

. ∵\because Area of smaller region bounded by

y2=4xy^2=4 x

and

x2+y2=5x^2+y^2=5
=2{area of OACO+ area of CABC}=2\{\text{area of } O A C O+\text{ area of } C A B C\}
=2[∫012xdx+∫155−x2dx=2[∣43x32∣01+(12x5−x2+52sin⁡−1x5)]15=2[(43−0)+(0+5π4)−(1+52sin⁡−115)]=2[13+5π4−52sin⁡−115]=23+5π2−5sin⁡−115=23+5cos⁡−1(15)=23+5sin⁡−1(25)\begin{aligned} & =2\left[\int\limits_0^1 2 \sqrt{x} d x+\int\limits_1^{\sqrt{5}} \sqrt{5-x^2} d x\right. \\ & =2\left[\left|\frac{4}{3} x^{\frac{3}{2}}\right|_0^1+\left(\frac{1}{2} x \sqrt{5-x^2}+\frac{5}{2} \sin ^{-1} \frac{x}{\sqrt{5}}\right)\right]_1^{\sqrt{5}} \\ & =2\left[\left(\frac{4}{3}-0\right)+\left(0+\frac{5 \pi}{4}\right)-\left(1+\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right)\right] \\ & =2\left[\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right]=\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1} \frac{1}{\sqrt{5}} \\ & =\frac{2}{3}+5 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right) \end{aligned}
Q86
One of the points of intersection of the curves y=1+3x−2x2y=1+3 x-2 x^2 and y=1xy=\dfrac{1}{x} is (12,2)\left(\dfrac{1}{2}, 2\right). Let the area of the region enclosed by these curves be 124(l5+m)−nlog⁡e(1+5)\dfrac{1}{24}(l \sqrt{5}+\mathrm{m})-\mathrm{n} \log _{\mathrm{e}}(1+\sqrt{5}), where l, m,n∈Nl, \mathrm{~m}, \mathrm{n} \in \mathbf{N}. Then l+m+nl+\mathrm{m}+\mathrm{n} is equal to
A 30
B 29
C 31
D 32
Correct Answer
Option A
Solution

Solving curves

y=1+3x−2x2 & y=1xy=1+3 x-2 x^2 ~\& ~y=\frac{1}{x}
2x3−3x2−x+1=0⇒(2x−1)(x2−x−1)=0⇒x=12,x=1±52\begin{aligned} & 2 x^3-3 x^2-x+1=0 \\ & \Rightarrow \quad(2 x-1)\left(x^2-x-1\right)=0 \\ & \Rightarrow \quad x=\frac{1}{2}, x=\frac{1 \pm \sqrt{5}}{2} \end{aligned}

Area

=∫121+52(1+3x−2x2−1xdx)= \int\limits_{{1 \over 2}}^{{{1 + \sqrt 5 } \over 2}} {\left( {1 + 3x - 2{x^2} - {1 \over x}dx} \right)}
=[x+3x22−2x3−ln⁡x]= \left[ {x + {{3{x^2}} \over 2} - {{2x} \over 3} - \ln x} \right]
=5+12+38(5+1)2−112(5+1)−ln⁡(5+12)−(12+38−112−ln⁡12)= {{\sqrt 5 + 1} \over 2} + {3 \over 8}{(\sqrt 5 + 1)^2} - {1 \over {12}}(\sqrt 5 + 1) - \ln \left( {{{\sqrt 5 + 1} \over 2}} \right) - \left( {{1 \over 2} + {3 \over 8} - {1 \over {12}} - \ln {1 \over 2}} \right)
=124[12(5+1)+9(5+1)2−2(5+1)−12−9+2]−ln⁡(5+12×2)]= {1 \over {24}}\left[ {12(\sqrt 5 + 1) + 9{{(\sqrt 5 + 1)}^2} - 2(\sqrt 5 + 1) - 12 - 9 + 2] - \ln \left( {{{\sqrt 5 + 1} \over 2} \times 2} \right)} \right]
=124[12(5+1)+9(6+25)−2(55+1+35(5+1)−19)]−ln⁡(5+1)=124[145+15]−ln⁡(5+1)∴l=14,m=15,n=1I+m+n=30\begin{aligned} = & \frac{1}{24}[12(\sqrt{5}+1)+9(6+2 \sqrt{5})- \\ & 2(5 \sqrt{5}+1+3 \sqrt{5}(\sqrt{5}+1)-19)]-\ln (\sqrt{5}+1) \\ = & \frac{1}{24}[14 \sqrt{5}+15]-\ln (\sqrt{5}+1) \\ \therefore \quad & \quad l=14, m=15, n=1 \\ & \quad I+m+n=30 \end{aligned}
Q87
The area (in sq. units) of the region described by {(x,y):y2≤2x, and y≥4x−1} \left\{(x, y): y^2 \leq 2 x \text{, and } y \geq 4 x-1\right\} is
A 932\dfrac{9}{32}
B 1112\dfrac{11}{12}
C 89\dfrac{8}{9}
D 1132\dfrac{11}{32}
Correct Answer
Option A
Solution
 Area =∫−121(y+14−y22)dy\text{ Area }=\int\limits_{-\frac{1}{2}}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y
=[y28+y4−y36]−121=(18+14−16)−(132−18+148)=524+796=2796=932\begin{aligned} & =\left[\frac{y^2}{8}+\frac{y}{4}-\frac{y^3}{6}\right]_{-\frac{1}{2}}^1 \\ & =\left(\frac{1}{8}+\frac{1}{4}-\frac{1}{6}\right)-\left(\frac{1}{32}-\frac{1}{8}+\frac{1}{48}\right) \\ & =\frac{5}{24}+\frac{7}{96} \\ & =\frac{27}{96} \\ & =\frac{9}{32} \end{aligned}
Q88
The area of the region in the first quadrant inside the circle x2+y2=8x^2+y^2=8 and outside the parabola y2=2xy^2=2 x is equal to :
A π2−13\dfrac{\pi}{2}-\dfrac{1}{3}
B π−13\pi-\dfrac{1}{3}
C π−23\pi-\dfrac{2}{3}
D π2−23\dfrac{\pi}{2}-\dfrac{2}{3}
Correct Answer
Option C
Solution

We have,

x2+y2=8x^2+y^2=8

and

y2=2xy^2=2 x

Area of shaded region

=∫02y22dy+∫2228−y2dy=16[y3]02+[y28−y2+4sin⁡−1(y22)]222=43+π−2=π−23\begin{aligned} & =\int\limits_0^2 \frac{y^2}{2} d y+\int\limits_2^{2 \sqrt{2}} \sqrt{8-y^2} d y \\ & =\frac{1}{6}\left[y^3\right]_0^2+\left[\frac{y}{2} \sqrt{8-y^2}+4 \sin ^{-1}\left(\frac{y}{2 \sqrt{2}}\right)\right]_2^{2 \sqrt{2}} \\ & =\frac{4}{3}+\pi-2=\pi-\frac{2}{3} \end{aligned}
Q89
The area enclosed between the curves y=x∣x∣y=x|x| and y=x−∣x∣y=x-|x| is :
A 83\dfrac{8}{3}
B 23\dfrac{2}{3}
C 43\dfrac{4}{3}
D 1
Correct Answer
Option C
Solution
y=x∣x∣y=x|x|

&

y=x−∣x∣y=x-|x|
y=x∣x∣={x2,x>0−x2,x<0y=x−∣x∣={0,x>02x,x<0\begin{aligned} & y=x|x|= \begin{cases}x^2, & x>0 \\ -x^2, & x<0\end{cases} \\ & y=x-|x|= \begin{cases}0, & x>0 \\ 2 x, & x<0\end{cases} \end{aligned}

Point of intersections are

(0,0)(0,0)

&

(−2,4)(-2,4)

.

∴ Area =∫−20(−x2−2x)dx=[−x33−2x22]−20=−[83−4]=43 sq. unit \begin{aligned} \therefore \quad & \text{ Area }=\int\limits_{-2}^0\left(-x^2-2 x\right) d x \\ & =\left[\frac{-x^3}{3}-\frac{2 x^2}{2}\right]_{-2}^0 \\ & =-\left[\frac{8}{3}-4\right]=\frac{4}{3} \text{ sq. unit } \end{aligned}
Q90
If the area of the region $$\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0
A 1
B 0
C 2
D −-1
Correct Answer
Option D
Solution
\left\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0

\Rightarrow \int\limits_1^2 {\left( {{1 \over x} - {a \over {{x^2}}}} \right)dx = \left| {\ln |x| + {a \over x}} \right|_1^2}

\begin{aligned} & \left(\ln 2+\frac{a}{2}\right)-(\ln 1+a)=\ln 2-\frac{a}{2} \\ & \Rightarrow \quad \frac{a}{2}=\frac{1}{7} \\ & \Rightarrow \quad a=\frac{2}{7} \\ & \quad 7 a-3=\frac{2}{7} \times 7-3=-1 \end{aligned}$$

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