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Area Under Curves

JEE Mathematics · 107 questions · Page 10 of 11 · Click an option or "Show Solution" to reveal answer

Q91
Let the area of the region enclosed by the curves y=3x,2y=27−3xy=3 x, 2 y=27-3 x and y=3x−xxy=3 x-x \sqrt{x} be AA. Then 10A10 A is equal to
A 172
B 154
C 162
D 184
Correct Answer
Option C
Solution
y′=3−32xy′=3(1−x2)\begin{aligned} & y^{\prime}=3-\frac{3}{2} \sqrt{x} \\ & y^{\prime}=3\left(1-\frac{\sqrt{x}}{2}\right) \end{aligned}
 Area =∫03(3x−(3x−x3/2))dx+∫3927−3x2−(3x−x3/2)dx=∫03x3/2dx+12∫39(27−9x+2x3/2)dx=[25x5/2]03+12[27x−9x22+2×25x5/2]39=25×93+12[243−7292+45×81×3−81+812−45×93]=12[486−729−812+9725]=815A=815∴10A=10×815=162\begin{aligned} & \text{ Area }=\int\limits_0^3\left(3 x-\left(3 x-x^{3 / 2}\right)\right) d x+\int\limits_3^9 \frac{27-3 x}{2}-\left(3 x-x^{3 / 2}\right) d x \\ & =\int\limits_0^3 x^{3 / 2} d x+\frac{1}{2} \int\limits_3^9\left(27-9 x+2 x^{3 / 2}\right) d x \\ & =\left[\frac{2}{5} x^{5 / 2}\right]_0^3+\frac{1}{2}\left[27 x-\frac{9 x^2}{2}+2 \times \frac{2}{5} x^{5 / 2}\right]_3^9 \\ & =\frac{2}{5} \times 9 \sqrt{3}+\frac{1}{2}\left[243-\frac{729}{2}+\frac{4}{5} \times 81 \times 3-81+\frac{81}{2}-\frac{4}{5} \times 9 \sqrt{3}\right] \\ & =\frac{1}{2}\left[\frac{486-729-81}{2}+\frac{972}{5}\right]=\frac{81}{5} \\ & A=\frac{81}{5} \\ & \therefore 10 A=10 \times \frac{81}{5}=162 \end{aligned}
Q92
Let the area enclosed between the curves ∣y∣=1−x2|y| = 1 - x^2 and x2+y2=1x^2 + y^2 = 1 be α\alpha. If 9α=βπ+γ;β,γ9\alpha = \beta \pi + \gamma; \beta, \gamma are integers, then the value of ∣β−γ∣|\beta - \gamma| equals:
A 15
B 18
C 33
D 27
Correct Answer
Option C
Solution
C1:∣y∣=1−x2C2:x2+y2=1\begin{aligned} & C_1:|y|=1-x^2 \\ & C_2: x^2+y^2=1 \end{aligned}

∴\therefore Required Area =α=4[=\alpha=4\left[\right. Area of circle in 1st 1^{\text{st }} quad.

−∫01(1−x2)dx]\left.-\int_0^1\left(1-\mathrm{x}^2\right) \mathrm{dx}\right]

=4[π4−[x−x33]01]α=π−83∴3α=3π−8∴9α=9π−24∴β=9,γ=−24∴∣β−γ∣=33\begin{aligned} & =4\left[\frac{\pi}{4}-\left[\mathrm{x}-\frac{\mathrm{x}^3}{3}\right]_0^1\right] \\ & \alpha=\pi-\frac{8}{3} \\ & \therefore 3 \alpha=3 \pi-8 \\ & \therefore 9 \alpha=9 \pi-24 \\ & \therefore \beta=9, \gamma=-24 \\ & \therefore|\beta-\gamma|=33 \end{aligned}
Q93
The area of the region enclosed by the curves y=x2−4x+4y=x^2-4 x+4 and y2=16−8xy^2=16-8 x is :
A 83\dfrac{8}{3}
B 55
C 88
D 43\dfrac{4}{3}
Correct Answer
Option A
Solution

Consider the curves

y=x2−4x+4=(x−2)2y = x^2 - 4x + 4 = (x-2)^2

and

y2=16−8x.y^2 = 16 - 8x.

Notice that the second equation can be rewritten in terms of

xx

:

8x=16−y2⟹x=2−y28.8x = 16 - y^2 \quad \Longrightarrow \quad x = 2 - \frac{y^2}{8}.

Step 1. Find the Intersection Points To find the points where the curves intersect, substitute

y=(x−2)2y = (x-2)^2

into

y2=16−8x.y^2 = 16 - 8x.

Let

u=x−2so thaty=u2.u = x - 2 \quad \text{so that} \quad y = u^2.

Then

y2=u4y^2 = u^4

and

x=u+2.x = u + 2.

Substitute into the second curve:

u4=16−8(u+2).u^4 = 16 - 8(u+2).

Simplify the right side:

16−8(u+2)=16−8u−16=−8u.16 - 8(u+2) = 16 - 8u - 16 = -8u.

Thus, the equation becomes

u4+8u=0⟹u(u4/u?  ∗∗correctingfactor∗∗).u^4 + 8u = 0 \quad \Longrightarrow \quad u(u^4/ u? \,\, **{correcting factor}**).

In fact, factor by taking out a common factor

uu

:

u(u3+8)=0.u(u^3 + 8) = 0.

Thus, either

u=0oru3=−8.u = 0 \quad \text{or} \quad u^3 = -8.

For

u=0u = 0

:

x=u+2=2,y=u2=0.x = u + 2 = 2, \quad y = u^2 = 0.

For

u3=−8u^3 = -8

:

u=−2,sox=−2+2=0,y=(−2)2=4.u = -2, \quad \text{so} \quad x = -2 + 2 = 0, \quad y = (-2)^2 = 4.

The curves intersect at the points

(2,0)(2, 0)

and

(0,4).(0, 4).

Step 2. Express the Curves in Terms of

yy

It is easier to integrate horizontally by expressing

xx

as a function of

yy

. From the second curve:

x=2−y28.x = 2 - \frac{y^2}{8}.

From the first curve, solving

y=(x−2)2y = (x-2)^2

for

xx

gives

x−2=±y.x-2 = \pm\sqrt{y}.

Since at the intersection

(0,4)(0, 4)

the

xx

–value is less than 2, we take the negative branch:

x=2−y.x = 2 - \sqrt{y}.

Thus, for a fixed

yy

between 0 and 4, the left boundary is

xleft=2−y,x_{\text{left}} = 2 - \sqrt{y},

and the right boundary is

xright=2−y28.x_{\text{right}} = 2 - \frac{y^2}{8}.

Step 3. Set Up the Integral for the Area The horizontal distance between the curves at a given

yy

is

Δx=xright−xleft=(2−y28)−(2−y)=y−y28.\Delta x = x_{\text{right}} - x_{\text{left}} = \left(2 - \frac{y^2}{8}\right) - \left(2 - \sqrt{y}\right) = \sqrt{y} - \frac{y^2}{8}.

Integrate with respect to

yy

from

y=0y = 0

to

y=4y = 4

:

A=∫04(y−y28) dy.A = \int_{0}^{4} \left(\sqrt{y} - \frac{y^2}{8}\right) \, dy.

Step 4. Evaluate the Integral Write the integral as

A=∫04y1/2 dy−18∫04y2 dy.A = \int_{0}^{4} y^{1/2}\, dy - \frac{1}{8} \int_{0}^{4} y^2\, dy.

Compute each term: For the first integral:

∫y1/2 dy=23y3/2.\int y^{1/2}\, dy = \frac{2}{3} y^{3/2}.

For the second integral:

∫y2 dy=y33.\int y^2\, dy = \frac{y^3}{3}.

Thus,

A=[23y3/2]04−18[y33]04.A = \left[\frac{2}{3} y^{3/2}\right]_0^4 - \frac{1}{8}\left[\frac{y^3}{3}\right]_0^4.

Substitute

y=4y = 4

(note that at

y=0y = 0

both terms vanish): Compute

43/24^{3/2}

:

43/2=(4)3=23=8.4^{3/2} = \left( \sqrt{4} \right)^3 = 2^3 = 8.

Compute the first term:

23×8=163.\frac{2}{3} \times 8 = \frac{16}{3}.

Compute the second term:

18⋅643=6424=83.\frac{1}{8} \cdot \frac{64}{3} = \frac{64}{24} = \frac{8}{3}.

Therefore, the area is

A=163−83=83.A = \frac{16}{3} - \frac{8}{3} = \frac{8}{3}.

Final Answer The area of the region enclosed by the curves is

83.\frac{8}{3}.
Q94
The area of the region bounded by the curves x(1+y2)=1x(1+y^2)=1 and y2=2xy^2=2x is:
A π4−13\dfrac{\pi}{4} - \dfrac{1}{3}
B π2−13\dfrac{\pi}{2} - \dfrac{1}{3}
C 2(π2−13)2\left(\dfrac{\pi}{2} - \dfrac{1}{3}\right)
D 12(π2−13)\dfrac{1}{2}\left(\dfrac{\pi}{2} - \dfrac{1}{3}\right)
Correct Answer
Option B
Solution
x(1+y2)=1..... (1)y2=2x..... (2)\begin{aligned} & x\left(1+y^2\right)=1 \quad\text{..... (1)}\\ & y^2=2 x\quad\text{..... (2)} \end{aligned}
\begin{aligned} &\text{ From equation (1) & (2) }\\ &\begin{aligned} x(1+2 x)=1 & \Rightarrow 2 x^2+x-1=0 \\ & \Rightarrow x=\frac{1}{2}, x=-1 \text{ (Reject) } \\ & \Rightarrow y^2=2\left(\frac{1}{2}\right) \\ & \Rightarrow y= \pm 1 \end{aligned} \end{aligned}
 Area bounded =∫−11(11+y2−y22)dy=(tan⁡−1y−y36)∣−11=π2−13\begin{aligned} \text{ Area bounded } & =\int_{-1}^1\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right) d y \\ & =\left.\left(\tan ^{-1} y-\frac{y^3}{6}\right)\right|_{-1} ^1 \\ & =\frac{\pi}{2}-\frac{1}{3} \end{aligned}
Q95
If the area of the region {(x,y):−1≤x≤1,0≤y≤a+e∣x∣−e−x,a>0}\left\{(x, y):-1 \leq x \leq 1,0 \leq y \leq \mathrm{a}+\mathrm{e}^{|x|}-\mathrm{e}^{-x}, \mathrm{a}>0\right\} is e2+8e+1e\dfrac{\mathrm{e}^2+8 \mathrm{e}+1}{\mathrm{e}}, then the value of aa is :
A 7
B 5
C 6
D 8
Correct Answer
Option B
Solution
 required area is a+∫01(a+ex−e−x)dxa+[a+ex+e−x]012a+e−1+e−1−1=e+8+1e2a=10⇒a=5\begin{aligned} & \text{ required area is } a+\int\limits_0^1\left(a+e^x-e^{-x}\right) d x \\ & a+\left[a+e^x+e^{-x}\right]_0^1 \\ & 2 a+e-1+e^{-1}-1=e+8+\frac{1}{e} \\ & 2 a=10 \Rightarrow a=5 \end{aligned}
Q96
The area of the region {(x,y):x2+4x+2≤y≤∣x+2∣}\left\{(x, y): x^2+4 x+2 \leq y \leq|x+2|\right\} is equal to
A 7
B 24/5
C 20/3
D 5
Correct Answer
Option C
Solution
x2+4x+2≤y≤∣x+2∣x^2+4 x+2 \leq y \leq|x+2|

The area bounded between

y=x2+4x+2=(x+2)2−2y=x^2+4 x+2=(x+2)^2-2

and y=∣x+2∣y=|x+2| is same as area bounded between y=x2−2y=x^2-2 and y=∣x∣y=|x| For P.O.I ∣x∣2−2=∣x∣|x|^2-2=|x|

⇒∣x∣=2⇒x=±2∴ Required area =−∫−22(x2−2)dx+∫−22∣x∣dx=−2∫02(x2−2)dx+2∫02x⋅dx=−2[x33−2x]02+2[x22]02=−−2[83−4]+2[42]=−2×(−43)+4=203\begin{aligned} & \Rightarrow|x|=2 \Rightarrow x= \pm 2 \\ & \therefore \text{ Required area }=-\int_{-2}^2\left(x^2-2\right) d x+\int_{-2}^2|x| d x \\ & =-2 \int_0^2\left(x^2-2\right) d x+2 \int_0^2 x \cdot d x \\ & =-2\left[\frac{x^3}{3}-2 x\right]_0^2+2\left[\frac{x^2}{2}\right]_0^2 \\ & =--2\left[\frac{8}{3}-4\right]+2\left[\frac{4}{2}\right] \\ & =-2 \times\left(\frac{-4}{3}\right)+4 \\ & =\frac{20}{3} \end{aligned}
Q97
The area of the region enclosed by the curves y=ex,y=∣ex−1∣y=\mathrm{e}^x, y=\left|\mathrm{e}^x-1\right| and yy-axis is :
A 1+log⁡e21+\log _{\mathrm{e}} 2
B log⁡e2\log _{\mathrm{e}} 2
C 1−log⁡e21-\log _{\mathrm{e}} 2
D 2log⁡e2−12 \log _{\mathrm{e}} 2-1
Correct Answer
Option C
Solution
 For Area ∫−ln⁡20[ex−(1−ex)]dx∫−nn⁡20(2ex−1)dx=[2ex−x]−ln⁡20=(2−(1+ℓn2))=1−ℓn2\begin{aligned} &\text{ For Area } \int_{-\ln 2}^0\left[e^x-\left(1-e^x\right)\right] d x\\ &\begin{aligned} & \int_{-\operatorname{nn} 2}^0\left(2 \mathrm{e}^{\mathrm{x}}-1\right) \mathrm{dx}=\left[2 \mathrm{e}^{\mathrm{x}}-\mathrm{x}\right]_{-\ln 2}^0 \\ & =(2-(1+\ell \mathrm{n} 2)) \\ & =1-\ell \mathrm{n} 2 \end{aligned} \end{aligned}
Q98
The area (in sq. units) of the region {(x,y):0≤y≤2∣x∣+1,0≤y≤x2+1,∣x∣≤3}\left\{(x, \mathrm{y}): 0 \leq \mathrm{y} \leq 2|x|+1,0 \leq \mathrm{y} \leq x^2+1,|x| \leq 3\right\} is
A 323\dfrac{32}{3}
B 643\dfrac{64}{3}
C 173\dfrac{17}{3}
D 803\dfrac{80}{3}
Correct Answer
Option B
Solution
 Area =2[∫02(x2+1)dx+∫23(2x+1)dx]⇒643∴(2)\begin{aligned} & \text{ Area }=2\left[\int_0^2\left(\mathrm{x}^2+1\right) \mathrm{dx}+\int_2^3(2 \mathrm{x}+1) \mathrm{dx}\right] \\ & \Rightarrow \frac{64}{3} \quad \therefore(2) \end{aligned}
Q99
If the area of the region {(x,y):1+x2≤y≤min⁡{x+7,11−3x}} \{(x, y) : 1 + x^2 \leq y \leq \min \{x+7, 11-3x\}\} is A A , then 3A 3A is equal to :
A 50
B 46
C 49
D 47
Correct Answer
Option A
Solution
A=∫−21(x+7−x2−1)dx+∫12(11+3x−x2−1)dx=[x22+6x−x33]−21+[10x−3x22−x33]12=503⇒3A=50 Option (1) \begin{aligned} & A=\int_{-2}^1\left(x+7-x^2-1\right) d x+\int_1^2\left(11+3 x-x^2-1\right) d x \\ & =\left[\frac{x^2}{2}+6 x-\frac{x^3}{3}\right]_{-2}^1+\left[10 x-\frac{3 x^2}{2}-\frac{x^3}{3}\right]_1^2 \\ & =\frac{50}{3} \Rightarrow 3 A=50 \quad \text{ Option (1) } \end{aligned}
Q100
Let f:[0,∞)→Rf:[0, \infty) \rightarrow \mathbb{R} be a differentiable function such that f(x)=1−2x+∫0xex−tf(t)dtf(x)=1-2 x+\int_0^x e^{x-t} f(t) d t for all x∈[0,∞)x \in[0, \infty). Then the area of the region bounded by y=f(x)y=f(x) and the coordinate axes is
A 5\sqrt5
B 2
C 2\sqrt2
D 12\dfrac{1}{2}
Correct Answer
Option D
Solution
∵f(x)=1−2x+∫0xex−tf(t)dt\because f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t

or, f(x)=1−2x+ex∫0xe−tf(t)dtf(x)=1-2 x+e^x \int_0^x e^{-t} f(t) d t on differentiating both sides w.r.t. xx we get

f′(x)=−2+ex∫0xe−tf(t)dt+ex⋅e−xf(x)f′(x)=−2+f(x)+2x−1+f(x){ from eq. (1) }\begin{aligned} & f^{\prime}(x)=-2+e^x \int_0^x e^{-t} f(t) d t+e^x \cdot e^{-x} f(x) \\ & f^{\prime}(x)=-2+f(x)+2 x-1+f(x)\{\text{ from eq. (1) \}} \end{aligned}
∴f(x)−2f(x)=2x−3 I.F. =e∫−2dx=e−2x∴e−2x⋅f(x)=∫e−2x(2x−3)dxe−2x⋅f(x)=(2x−3)⋅e−2x−2−∫2⋅e−2x−2dxe−2x⋅f(x)=(2x−3)e−2x−2+e−2x−2+cf(x)=−x+1+c′e2x∵f(x)=1 from eq. (1) ∴1=0+1+c′⇒c′=0∴f(x)=−x+1\begin{aligned} & \therefore \quad f(x)-2 f(x)=2 x-3 \\ & \text{ I.F. }=e^{\int-2 d x}=e^{-2 x} \\ & \therefore \quad e^{-2 x} \cdot f(x)=\int e^{-2 x}(2 x-3) d x \\ & e^{-2 x} \cdot f(x)=(2 x-3) \cdot \frac{e^{-2 x}}{-2}-\int 2 \cdot \frac{e^{-2 x}}{-2} d x \\ & e^{-2 x} \cdot f(x)=\frac{(2 x-3) e^{-2 x}}{-2}+\frac{e^{-2 x}}{-2}+c \\ & f(x)=-x+1+c^{\prime} e^{2 x} \\ & \because \quad f(x)=1 \text{ from eq. (1) } \\ & \therefore \quad 1=0+1+c^{\prime} \Rightarrow c^{\prime}=0 \\ & \therefore \quad f(x)=-x+1 \end{aligned}

⇒\Rightarrow \quad Area =12=\dfrac{1}{2}

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