Application of Integrals Test

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Application of Integrals Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

The area of the figure bounded by $$f\left(x\right)=\sin{x}, g\left(x\right)=\cos{x}$$ in the first quadrant is:

A
$$2\left(\sqrt{2}-1\right)$$ sq.unit

B
$$\sqrt{3}+1$$ sq.unit

C
$$2\left(\sqrt{3}-1\right)$$.sq.unit

D
none of these.

Solution

We know that $$\sin{x}\ge \cos{x}$$ for $$x\in\left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right]$$
and $$\sin{x}\le \cos{x}$$ for $$x\in\left[0,\dfrac{\pi}{4}\right]$$
$$\therefore$$ the required area is
$$=\int_{0}^{\frac{\pi}{4}}{\left(\cos{x}-\sin{x}\right)dx}+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\left(\sin{x}-\cos{x}\right)dx}$$
$$=\left(\sin{x}+\cos{x}\right)_{0}^{\frac{\pi}{4}}+\left(\sin{x}+\cos{x}\right)_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$
$$=\left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)$$
$$=2\left(\sqrt{2}-1\right)$$.sq.unit.
Question 2
1 / -0

Points of inflexion of the curve
$$y = x^4 - 6x^3 + 12x^2 + 5x + 7$$ are

A
$$(1, 19); (1, 12)$$

B
$$(1, 19); (2, 33)$$

C
$$(1, 2); (2, 1)$$

D
$$(1, 7); (2, 6)$$

Solution

$$y=x^{4}-6x^{3}+12x^{2}+5x+7$$
$$y(x)=4x^{3}-18x^{2}+24x+5$$
$$y(x)=12x^{2}-36x+24$$
$$y(x)=0$$
$$12x^{2}-36x+24=0$$
$$x^{2}-3x+2=0$$
$$x^{2}-2x-x+2=0$$
$$x(x-2)-1(x-2)=0$$
$$(x-1)-1(x-2)=0$$
$$(x-1)(x-2)=0$$
$$x=1, 2$$
Inflection point of a function is where the function changes from concave up to concave down or vice-versa
$$x<1, f(x)>0$$
$$1<x<2, f(x)<0$$
$$x>2, f(x)>0$$
$$\because f(x)$$ changes sign
$$\therefore$$ At $$x=1, 2y=f(x)$$ has inflection point
At $$x=1, y=f(1)=19$$
At $$x=2, y=f(2)=33$$
Point of inflection $$(1,19); (2,33)$$
Question 3
1 / -0

The area under the curve $$y=2x^3+4x^2$$ between $$x=2,x=4$$ is

A
$$192.6$$

B
$$198.6$$

C
$$88.3$$

D
$$172.3$$

Solution

The area under the curve is given as $$\displaystyle \int _2^4 2x^3+4x^2 dx\\\displaystyle \int _2^4 2x^3dx+\int _2^4 4x^2dx\\\left.2\dfrac{x^4}{4}\right|_2^4+\left.4\dfrac{x^3}{3}\right|_2^4\\64\times 2-8+\dfrac{4^4}3-\dfrac{32}{3}\\120- \dfrac{224}{3}=192.6$$
Question 4
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If the curves $$y=x^3+ax$$ and $$y=bx^2+c$$ pass through the point $$(-1, 0)$$ and have common tangent line at this point, then the value of $$a+b$$ is?

A
$$0$$

B
$$-2$$

C
$$-3$$

D
$$-1$$

Solution

As the curve pass through the point $$P(-1,0)$$
$$0 = -1-a$$
$$\implies a = -1$$
$$0 = b+c$$
Common tangent at this point
$$\cfrac{dy}{dx} = 2bx$$ and $$\cfrac{dy}{dx} = 3x^2+a = 3-1 = 2$$
$$2bx = 2$$
$$-2b = 2$$
$$b = -1$$
$$a+b = -2$$
Question 5
1 / -0

The area (in sq. units) of the region $$\{ x \in R:x \ge ,y \ge 0,y \ge x - 2\ $$ and $$y \le \sqrt x \} $$, is

A
$$\dfrac{{13}}{3}$$

B
$$\dfrac{{8}}{3}$$

C
$$\dfrac{{16}}{3}$$

D
$$\dfrac{{5}}{3}$$

Solution

The given reqion $$\{x\in R: x\geq 0, y\geq 0, y\geq x-2 and y\leq \sqrt{x}\}$$

Here $$y\geq x-2$$ and $$y\leq \sqrt{x}$$

$$\Rightarrow x-2\leq y\leq \sqrt{x}$$

Suppose

$$\Rightarrow y=x-2$$ …….$$(1)$$ and $$y=\sqrt{x}$$ …………$$(2)$$

From $$(1)$$ and $$(2)$$

$$\Rightarrow x-2=\sqrt{x}$$

$$\Rightarrow (x-2)^2=(\sqrt{x})^2$$ [squaring both sides]

$$\Rightarrow x^2+4-4x=x$$

$$\Rightarrow x^2-4x+4-x=0$$

$$\Rightarrow x(x-4)-1(x-4)=0$$

$$\Rightarrow (x-1)(x-4)=0$$

$$\therefore x=1, 4$$

At $$x=1$$, from $$(1)$$, $$y=1-2=-1$$ but $$y\geq 0$$

$$\therefore x=1$$ is neglected

At $$x=4$$, from $$(1)$$, $$y=4-2=2$$

$$\therefore$$ Area of the region $$=\displaystyle\int^4_0\{\sqrt{x}-(x-2)\}dx=\displaystyle\int^4_0(x^{\dfrac{1}{2}}-x+2)dx$$

$$=\left[\dfrac{x^{\dfrac 12+1}}{\dfrac 12+1}-\dfrac{x^2}{\alpha}+2x\right]^4_0$$

$$=\dfrac{4^{3/2}}{\dfrac 32}-\dfrac{4^2}{2}+2(4)-0+0-0$$

$$=\dfrac{2}{3}\times 8-8+8=\dfrac{16}{3}$$.
Question 6
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The area of the plane region bounded by the curves $$x + 2 y ^ { 2 }= 0 \text { and } x + 3 y ^ { 2 } = 1$$

A
$$\dfrac { 4 } { 3 }$$

B
$$\dfrac { 5 } { 3 }$$

C
$$\dfrac { 2 } { 3 }$$

D
$$\dfrac { 1 } { 3 }$$

Solution

$$\begin{array}{l} 1-3{ y^{ 2 } }=-2{ y^{ 2 } }\Rightarrow { y^{ 2 } }=1\, \therefore y=\pm 1 \\ y=-1\, \, \Rightarrow x=-2 \\ y=1\, \, \, \, \Rightarrow x=-2 \\ The\, \, bounded\, \, region\, \, is\, as\, under \\ The\, \, desired\, \, area\, \, =2\int _{ 0 }^{ 1 }{ \left[ { \left( { 1-3{ y^{ 2 } } } \right) -\left( { -2{ y^{ 2 } } } \right) } \right] }dy \\ =22\int _{ 0 }^{ 1 }{ \left[ { \left( { 1-{ y^{ 2 } } } \right) } \right] }dy=2\left[ { y-\cfrac { { { y^{ 3 } } } }{ 3 } } \right] _{ 0 }^{ 1 } \\ =2\times \cfrac { 2 }{ 3 } =\cfrac { 4 }{ 3 } sq.\, units \end{array}$$

Hence, this is the answer.
Question 7
1 / -0

What is the area of the region enclosed between the curve $$y^2=2x$$ and the straight line $$y=x$$ ?

A
$$\dfrac{2}{3}$$ square units

B
$$\dfrac{4}{3}$$ square units

C
$$\dfrac{1}{3}$$ square units

D
$$1$$ square unit

Solution

Given,

$$y^2=2x$$

$$y=x$$

$$\therefore x^2=2x$$

$$x=2$$

area $$=\int_{0}^{2}(\sqrt{2x})-x$$

$$=\sqrt{2}\left [ \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_0^2 - \left [ \dfrac{x^2}{2} \right ]_0^2$$

$$=\sqrt{2}\dfrac{4\sqrt{2}}{3}-\dfrac{2^2}{2}$$

$$=\dfrac{8}{3}-2$$

$$=\dfrac{2}{3}$$sq.units
Question 8
1 / -0

The area bounded by the parabola $$y=x^{2}$$ and the straight line $$\mathrm{y}=2\mathrm{x}$$ is

A
$$\displaystyle \frac{4}{3}$$ sq. units

B
$$\displaystyle \frac{3}{4}$$ sq. units

C
$$\displaystyle \frac{2}{3}$$ sq. units

D
$$\displaystyle \frac{1}{3}$$ sq. units

Solution

$$\displaystyle \int _{ 0 }^{ 2 }{ \left( 2x-{ x }^{ 2 } \right) dx } ={ \left[ \frac { 2{ x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2 }=\frac { 4 }{ 3 } $$
Question 9
1 / -0

The area bounded by the two parabolas $$y^{2}=8x$$ and $$x^{2}=8y$$ is

A
$$64$$ sq. units

B
$$\displaystyle \frac{64}{3}$$ sq, units

C
$$\displaystyle \frac{32}{3}$$ sq. units

D
$$\displaystyle \frac{1}{3}$$ sq. units

Solution

$$y^{2}=8x$$ and $$x^{2}=8y$$
the points of intersection
$$x^2/8=\sqrt{8x}$$
$$\dfrac{x^3}{8^2}=8$$
x=0 and x=8
So, area b//w curves is
$$\overset{8 }{\underset{0}{\int}}(\sqrt{8x}-x^2/8)dx$$
$$\sqrt{8}\overset{8 }{\underset{0}{\int}}\sqrt{x}dx-\overset{8 }{\underset{0}{\int}}\left ( \dfrac{x^3}{8} \right )dx=\dfrac{2\sqrt{8}}{3}(x)^{3/2}|_{0}^{8}-1/8\dfrac{x^3}{3}|_{0}^{8}$$
$$2/3 8^2-1/8(8^3/3)$$
$$=\dfrac{2.8^2}{3}-8^2/3=\dfrac{64}{3} sq\: units$$
Question 10
1 / -0

The area of the region bounded by the curve $$y=x^{2}+1$$ and $$y=2x-2$$ between $${x}=-1$$ and $${x}=2$$ is:

A
$$9$$sq. units

B
$$12$$sq. units

C
$$15$$sq. units

D
$$14$$sq. units

Solution

We have to find area of the region bounded by curves $$y=x^{2}+1$$ and $$y=2x-2$$ between $$\:x=-1\: and\: x=2$$
To find points of intersections, if any, for the parabola and the straight line we solve both simultaneously.
$$x^2+1=2x-2$$
$$ \Rightarrow x^2-2x+3=0$$, which has no real solutions. Hence, no points of intersection for the parabola and the straight line.
From the figure, the graph of $$y=x^2+1$$ will be always above the graph of $$y=2x-2$$.
Hence the required area is $$\overset{2 }{\underset{-1}{\int}}[(x^{2}+1)-(2x-2)]dx$$

$$\overset{2 }{\underset{-1}{\int}}(x^2-2x+3)dx$$

$$=\dfrac{x^3}{3}|_{-1}^{2}-x^2|_{-1}^{2}+3x|_{-1}^{2}$$

$$=3-(3)+9$$

$$=9\: sq \:units.$$

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Application of Integrals Test - 13

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Application of Integrals Test - 13

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SHARING IS CARING

Question 1

$$2\left(\sqrt{2}-1\right)$$ sq.unit

$$\sqrt{3}+1$$ sq.unit

$$2\left(\sqrt{3}-1\right)$$.sq.unit

none of these.

Solution

Question 2

$$(1, 19); (1, 12)$$

$$(1, 19); (2, 33)$$

$$(1, 2); (2, 1)$$

$$(1, 7); (2, 6)$$

Solution

Question 3

$$192.6$$

$$198.6$$

$$88.3$$

$$172.3$$

Solution

Question 4

$$0$$

$$-2$$

$$-3$$

$$-1$$

Solution

Question 5

$$\dfrac{{13}}{3}$$

$$\dfrac{{8}}{3}$$

$$\dfrac{{16}}{3}$$

$$\dfrac{{5}}{3}$$

Solution

Question 6

$$\dfrac { 4 } { 3 }$$

$$\dfrac { 5 } { 3 }$$

$$\dfrac { 2 } { 3 }$$

$$\dfrac { 1 } { 3 }$$

Solution

Question 7

$$\dfrac{2}{3}$$ square units

$$\dfrac{4}{3}$$ square units

$$\dfrac{1}{3}$$ square units

$$1$$ square unit

Solution

Question 8

$$\displaystyle \frac{4}{3}$$ sq. units

$$\displaystyle \frac{3}{4}$$ sq. units

$$\displaystyle \frac{2}{3}$$ sq. units

$$\displaystyle \frac{1}{3}$$ sq. units

Solution

Question 9

$$64$$ sq. units

$$\displaystyle \frac{64}{3}$$ sq, units

$$\displaystyle \frac{32}{3}$$ sq. units

$$\displaystyle \frac{1}{3}$$ sq. units

Solution

Question 10

$$9$$sq. units

$$12$$sq. units

$$15$$sq. units

$$14$$sq. units

Solution

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