Area Under Curves — Page 8 | JEE Mathematics

Q71

The area enclosed by the curves

{y^2} + 4x = 4

and

y - 2x = 2

is :

A

{{22} \over 3}

B 9

C

{{23} \over 3}

D

{{25} \over 3}

Correct Answer

Option B

Solution

Required area $=\int_{-4}^{2}\left(\dfrac{4-y^{2}}{4}-\dfrac{y-2}{2}\right) d y$ $=\int_{-4}^{2} \dfrac{8-2 y-y^{2}}{4} d y$ $=\dfrac{1}{4}\left\{8 y-y^{2}-\dfrac{y^{3}}{3}\right\}_{-4}^{2}$ $=9$ square units

Q72

The area of the region

\left\{(x, y): x^{2} \leq y \leq\left|x^{2}-4\right|, y \geq 1\right\}

is

A

\dfrac{4}{3}(4 \sqrt{2}+1)

B

\dfrac{3}{4}(4 \sqrt{2}+1)

C

\dfrac{4}{3}(4 \sqrt{2}-1)

D

\dfrac{3}{4}(4 \sqrt{2}-1)

Correct Answer

Option C

Solution

\begin{aligned} & \text{ Required area }=2\left[\int_1^2 \sqrt{y} d y+\int_4^2 \sqrt{4-y} d y\right] \\\\ & \left.\left.=2\left[\frac{y^{3 / 2}}{\frac{3}{2}}\right]_1^2-\frac{2(4-y)^{3 / 2}}{3}\right]_2^4\right] \\\\ & =\frac{4}{3}(4 \sqrt{2}-1) \end{aligned}

Q73

The area of the region enclosed by the curve

f(x)=\max \{\sin x, \cos x\},-\pi \leq x \leq \pi

and the

x

-axis is

A

2 \sqrt{2}(\sqrt{2}+1)

B 4

C

2(\sqrt{2}+1)

D

4(\sqrt{2})

Correct Answer

Option B

Solution

Area = =

\begin{aligned} &\left|\int_{-\pi}^{-3 \pi / 4} \sin x d x\right|+\left|\int_{\pi / 4}^\pi \sin x d x\right|+\left|\int_{-3 \pi / 4}^{-\pi / 2} \cos x d x\right| +\left|\int_{-\pi / 2}^{\pi / 4} \cos x d x\right| \end{aligned}

\begin{aligned} & =\left|\frac{1}{\sqrt{2}}-1\right|+\left|1+\frac{1}{\sqrt{2}}\right|+\left|-1+\frac{1}{\sqrt{2}}\right|+\left|\frac{1}{\sqrt{2}}+1\right| \\\\ & = 2+\frac{2}{\sqrt{2}}+2-\frac{2}{\sqrt{2}}=4 \end{aligned}

Q74

If the area of the region bounded by the curves

y=4-\dfrac{x^2}{4}

and

y=\dfrac{x-4}{2}

is equal to

\alpha

, then

6 \alpha

. equals

A 210

B 250

C 240

D 220

Correct Answer

Option B

Solution

$y=4-\dfrac{x^2}{4}$ and $y=\dfrac{x-4}{2}$

\begin{aligned} & \text{ Area }=\int\limits_{-6}^4\left(4-\frac{x^2}{4}-\frac{x}{2}+2\right) d x \\ & =\left[6 x-\frac{x^3}{12}-\frac{x^2}{4}\right]_{-6}^4 \\ & =6(4+6)-\left(\frac{64}{12}+\frac{216}{12}\right)-\left(\frac{16}{4}-\frac{36}{4}\right) \\ & =60-\frac{70}{3}+5 \\ & \alpha=\frac{125}{3} \\ & 6 \alpha=6 \times \frac{125}{3}=250 \end{aligned}

Q75

The area of the region enclosed by the curve

y=x^{3}

and its tangent at the point

(-1,-1)

is :

A

\dfrac{23}{4}

B

\dfrac{19}{4}

C

\dfrac{27}{4}

D

\dfrac{31}{4}

Correct Answer

Option C

Solution

Given $y=x^3$

\begin{aligned} & \Rightarrow \frac{d y}{d x}=3 x^2 \\\\ & \left(\frac{d y}{d x}\right)_{(-1,-1)}=3 \end{aligned}

Equation of tangent at $(-1,-1)$

\begin{aligned} & (y+1)=3(x+1) \\\\ & y=3 x+2 \end{aligned}

Solving (i) and (ii)

x^3=3 x+2

We can rewrite this as $x^3 - 3x - 2 = 0$ .

This equation will give us the x-coordinates of the points where the line $y = 3x + 2$ intersects the curve $y = x^3$ .

We already know that one point of intersection is $(-1, -1)$ (as it's a point of tangency).

To find the other points of intersection, we can solve the cubic equation $x^3 - 3x - 2 = 0$ .

Let's factor this equation: $x^3 - 3x - 2 = 0 \Rightarrow (x + 1)(x + 1)(x - 2) = 0$ .

Setting each factor equal to zero gives the solutions $x = -1, -1, 2$ .

$\therefore$ Other point is $Q(2,8)$ .

We have the required area as

\int_{-1}^{2} (3x + 2 - x^3) \, dx = \left[ \frac{3}{2}x^2 + 2x - \frac{1}{4}x^4 \right]_{-1}^{2}.

Evaluating this at $x = 2$ and $x = -1$ gives

\left[ \frac{3}{2} \cdot (2)^2 + 2 \cdot 2 - \frac{1}{4} \cdot (2)^4 \right] - \left[ \frac{3}{2} \cdot (-1)^2 + 2 \cdot (-1) - \frac{1}{4} \cdot (-1)^4 \right]

= \left[ 6 + 4 - 4 \right] - \left[ \frac{3}{2} - 2 - \frac{1}{4} \right]

= 6 - \left( \frac{1}{4} \right)

= \frac{27}{4}

Q76

Area of the region

\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq 2 y\right\}

is

A

2 \pi+\dfrac{16}{3}

B

\pi-\dfrac{8}{3}

C

\pi+\dfrac{8}{3}

D

2 \pi-\dfrac{16}{3}

Correct Answer

Option D

Solution

We have,

x^2+(y-2)^2 \leq 2^2 \text{ and } x^2 \geq 2 y

On solving the given equation of parabola and circle, we get

\begin{aligned} 2 y+(y-2)^2 & =4 \\\\ \Rightarrow y =0 \text{ or } 2 \end{aligned}

If $y=0$ , then $x=0$ and If $y=2$ , then $x= \pm 2$ $\therefore$ The intersecting point of circle and parabola are $(0,0),(-2,2),(2,2)$ Area of the shaded region =

=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^2=4-\pi

\begin{aligned} \text{ Required area } & =2\left[\int\limits_0^2 \frac{x^2}{2} d x-(4-\pi)\right] \\\\ & =2\left[\left.\frac{x^3}{6}\right|_0 ^2-4+\pi\right] \\\\ & =2\left[\frac{4}{3}+\pi-4\right] \\\\ & =2\left[\pi-\frac{8}{3}\right] \\\\ & =2 \pi-\frac{16}{6} \end{aligned}

Q77

The area of the region

\left\{(x, y): x^{2} \leq y \leq 8-x^{2}, y \leq 7\right\}

is :

A 18

B 24

C 20

D 21

Correct Answer

Option C

Solution

The given curves are $x^2 \leq y, y \leq 8-x^2 ; y \leq 7$ On solving, we get $x^2=8-x^2$

\begin{aligned} & \Rightarrow x^2=4 \\\\ & \Rightarrow x= \pm 2 \end{aligned}

\text{ So, area }=2\left[\int_0^4 \sqrt{y} d y+\int_4^7 \sqrt{8-y} d y\right]

=2\left\{\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^4+\left[\frac{-(8-y)^{\frac{3}{2}}}{\frac{3}{2}}\right]_4^7\right\}

\begin{aligned} & =2 \times \frac{2}{3}\left\{\left[4^{3 / 2}-0\right]+\left(-(1)^{3 / 2}+(4)^{3 / 2}\right)\right\} \\\\ & =\frac{4}{3}\{8-1+8\}=\frac{4}{3} \times 15=20 \text{ sq. units } \end{aligned}

Q78

The area bounded by the curves

y=|x-1|+|x-2|

and

y=3

is equal to :

A 5

B 4

C 6

D 3

Correct Answer

Option B

Solution

Given equation of curve $y=|x-1|+|x-2|$ and $y=3$ When, $x-1=0 \Rightarrow x=1$ , then $y=1$ and when, $x-2=0 \Rightarrow x=2$ , then $y=1$ Equation of curve passing through point $(1,1)$ and $(2,1)$ and $(0,3),(3,3)$ $\therefore$ Area bounded by given curves

=\frac{1}{2}(1+3) \times 2(\text{ Area of trapezium })=4

Q79

The area enclosed by the curves

x y+4 y=16

and

x+y=6

is equal to :

A

28-30 \log _{\mathrm{e}} 2

B

30-28 \log _{\mathrm{e}} 2

C

30-32 \log _{\mathrm{e}} 2

D

32-30 \log _{\mathrm{e}} 2

Correct Answer

Option C

Solution

To find the enclosed area between the two curves $x y+4 y=16$ and $x+y=6$ , we need to determine the region of intersection and integrate the difference of the functions over the interval where they intersect.

First, let's solve the equations simultaneously to find the points of intersection.

The second curve $x+y=6$ can be rewritten as $y=6-x$ .

Substitute $y=6-x$ into the first equation: $x(6-x) + 4(6-x) = 16$ $6x - x^2 + 24 - 4x = 16$ $-x^2 + 2x + 8 = 0$ This can be simplified to: $x^2 - 2x - 8 = 0$ Using the quadratic formula, we can find the roots of this equation: $x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}$ $x = \dfrac{2 \pm \sqrt{4 + 32}}{2}$ $x = \dfrac{2 \pm \sqrt{36}}{2}$ $x = \dfrac{2 \pm 6}{2}$ So we have two solutions: $x_1 = \dfrac{2 + 6}{2} = 4$ $x_2 = \dfrac{2 - 6}{2} = -2$ For $x_1 = 4$ , substitute this back into the line equation $x+y=6$ : $4+y=6$ $y=2$ For $x_2 = -2$ , do the same: $-2+y=6$ $y=8$ So we have two points of intersection: $(4,2)$ and $(-2,8)$ .

Now, to find the area between $x y + 4y = 16$ and $x + y = 6$ , we'll integrate the top function minus the bottom function from $x = -2$ to $x = 4$ .

First, we need to express $y$ from both equations in terms of $x$ : For $x y + 4 y = 16$ , express $y$ in terms of $x$ : $y(x+4) = 16$ $y = \dfrac{16}{x+4}$ For $x + y = 6$ , we have: $y = 6 - x$ The area $A$ is therefore given by: $A = \int\limits_{-2}^{4}\left( 6 - x - \dfrac{16}{x+4} \right) dx$ Now, we integrate: $A = \int\limits_{-2}^{4} (6 - x) dx - \int_{-2}^{4} \dfrac{16}{x + 4} dx$ $A = \left[6x - \dfrac{x^2}{2}\right]_{-2}^{4} - \left[16 \ln|x + 4|\right]_{-2}^{4}$ $A = \left( 6 \times 4 - \dfrac{1}{2} \times 4^2 \right) - \left( 6 \times -2 - \dfrac{1}{2} \times -2^2 \right) - 16 \left( \ln|4+4| - \ln|-2+4| \right)$ $A = \left( 24 - 8 \right) - \left( -12 - 2 \right) - 16 \left( \ln 8 - \ln 2 \right)$ $A = 16 + 14 - 16 \left( \ln (2^3) - \ln 2 \right)$ $A = 30 - 16 \left( 3 \ln 2 - \ln 2 \right)$ $A = 30 - 16 (2 \ln 2)$ $A = 30- 32 \ln 2$

Q80

The area of the region enclosed by the parabolas

y=4 x-x^2

and

3 y=(x-4)^2

is equal to :

A

\dfrac{32}{9}

B

\dfrac{14}{3}

C 4

D 6

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Area }=\left\lvert\, \int\limits_1^4\left[\left(4 x-x^2\right)-\frac{(x-4)^2}{3}\right] d x\right. \\ & \text{ Area }=\left|\frac{4 x^2}{2}-\frac{x^3}{3}-\frac{(x-4)^3}{9}\right|_1^4 \\ & =\left|\left(\frac{64}{2}-\frac{64}{3}-\frac{4}{2}+\frac{1}{3}-\frac{27}{9}\right)\right| \\ & \Rightarrow(27-21)=6 \end{aligned}