Application of Integrals Test

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Application of Integrals Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The area of a loop bounded by the curve $$y=a \sin x$$ and x- axis is

A
$$a$$

B
$$2a^2$$

C
$$0$$

D
$$2a$$

Solution
Question 2
1 / -0

The area bounded by the curves $$\left|x\right|+\left|y\right|\ge 1$$ and $${x}^{2}+{y}^{2}\le 1$$ is:

A
$$2$$ sq.unit

B
$$\pi$$ sq.unit

C
$$\left(\pi-2\right)$$ sq.unit

D
$$\left(\pi+2\right)$$ sq.unit

Solution

From the figure shown
$$\therefore$$ Required Area$$=$$area of the circle$$-$$area of the square.
$$=\pi{\left(1\right)}^{2}-{\left(\sqrt{2}\right)}^{2}$$
$$=\left(\pi-2\right)$$.sq.unit
Question 3
1 / -0

The area bounded by the curves $$y=\left|x\right|-1$$ and $$y=-\left|x\right|+1$$ is

A
$$1$$.sq.unit

B
$$2$$.sq.unit

C
$$2\sqrt{2}$$.sq.unit

D
$$4\sqrt{2}$$.sq.unit

Solution

Required area$$=$$shaded area$$={\left(\sqrt{2}\right)}^{2}=2$$.sq.unit.
Question 4
1 / -0

Let $$f\left(x\right)={x}^{2}-3x+2$$ be a function,for all $$x\in R$$. On the basis of given information, answer the given question
The number of solutions of $$\left|y\right|=\left|f\left(\left|x\right|\right)\right|$$ and $${x}^{2}+{y}^{2}=2$$ is,

A
$$4$$

B
$$6$$

C
$$8$$

D
$$5$$

Solution
Question 5
1 / -0

Let $$f\left(x\right)={x}^{2}-3x+2$$ be a function, for all $$x\in R$$. On the basis of given information, answer the given question.
The area bounded by $$f\left(x\right),$$ the $$x-$$axis and $$y-$$axis is,

A
$$\dfrac{1}{3}$$.sq.unit

B
$$\dfrac{2}{3}$$.sq.unit

C
$$\dfrac{3}{5}$$.sq.unit

D
$$\dfrac{5}{6}$$.sq.unit

Solution
Question 6
1 / -0

The area of the figure bounded by two branches of the curve $${\left(y-x\right)}^{2}={x}^{3}$$ and the straight line $$x=1$$ is:

A
$$\dfrac{1}{3}$$ sq.unit

B
$$\dfrac{4}{5}$$ sq.unit

C
$$\dfrac{5}{4}$$ sq.unit

D
$$3$$ sq.unit

Solution

Given curve is $${\left(y-x\right)}^{2}={x}^{3}$$
$$\Rightarrow y-x=\pm x\sqrt{x}$$
or $$y=x\pm x\sqrt{x}$$
$$\therefore$$ Required Area$$=\left|\int_{0}^{1}{\left[(x+x\sqrt{x}\right)-\left(x-x\sqrt{x})\right]dx}\right|$$
$$=\left|2\int_{0}^{1}{{x}^{\frac{3}{2}}dx}\right|$$
$$=2\left[\dfrac{{x}^{\frac{5}{2}}}{\frac{5}{2}}\right]_{0}^{1}$$
$$=\left|2.\dfrac{2}{5}\right|=\dfrac{4}{5}$$.sq.unit.
Question 7
1 / -0

The area bounded by $$y=2-\left|2-x\right| , y=\dfrac{3}{\left|x\right|}$$ is

A
$$\dfrac{5-4\ln{2}}{3}$$.sq.unit

B
$$\dfrac{2-\ln{3}}{2}$$.sq.unit

C
$$\dfrac{4-3\ln{3}}{2}$$.sq.unit

D
none of these

Solution

$$y=2-\left|2-x\right|, y=\dfrac{3}{\left|x\right|}$$
$$y=\begin{cases}x; x\le 2 \\ 4-x; x\ge 2\end{cases}$$ and
$$y=\begin{cases} \dfrac{3}{x}; x>0 \\-\dfrac{3}{x}; x<0 \end{cases}$$
Hence,Required Area$$PQRSP=$$area$$PQRP+$$area$$PRSP$$
$$=\left|\int_{\sqrt{3}}^{2}{\left(x-\dfrac{3}{x}\right)dx}\right|+\left|\int_{2}^{3}\left[\left(4-x\right)-\dfrac{3}{x}\right]dx\right|$$
$$=\left[\dfrac{{x}^{2}}{2}-3\ln{x}\right]_{\sqrt{3}}^{2}+\left[4x-\dfrac{{x}^{2}}{2}-3\ln{x}\right]_{2}^{3}$$
On simplification,we get
$$=\dfrac{4-3\ln{3}}{2}$$.sq.unit
Question 8
1 / -0

If $$f\left(x+y\right)=f\left(x\right)+f\left(y\right)-xy$$ for all $$x,y\in R$$ and $$\lim _{ h\rightarrow 0 }{ \frac { f(h) }{ h } } =3$$, then the area bounded by the curves $$y=f\left(x\right)$$ and $$y={x}^{2}$$ is:

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Put $$x=y=0$$
$$\Rightarrow f\left(0\right)=0$$
Now,
$${f}{'}\left(x\right)=\displaystyle \lim_{h\rightarrow 0}{\dfrac{f\left(x+h\right)-f\left(x\right)}{h}}$$
$$=\displaystyle \lim_{h\rightarrow 0}{\dfrac{f\left(x\right)+f\left(h\right)-hx-f\left(x\right)}{h}}$$
$$=\displaystyle \lim_{h\rightarrow 0}{\dfrac{f\left(h\right)}{h}}-x=3-x$$
$$\therefore f\left(x\right)=3x-\dfrac{{x}^{2}}{2}+c$$ and $$f\left(0\right)=0$$
$$\Rightarrow c=0$$
$$\therefore f\left(x\right)=3x-\dfrac{{x}^{2}}{2}$$
Required Area$$=\left|\int_{0}^{2}{\left(3x-\dfrac{{x}^{2}}{2}-{x}^{2}\right)}dx\right|$$
$$=\left(\dfrac{3{x}^{2}}{2}-\dfrac{3{x}^{3}}{6}\right)_{0}^{2}$$
$$=6-4=2$$
Question 9
1 / -0

The area bounded by the curve $$y={(x-1)}^{2},\ ={(x+1)}^{2}$$ and the $$x-axis$$ is

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

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

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

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

Solution

$$A=2\int _{ 0 }^{ 1 }{ (x-1) }^{ 2 }dx\\ A={ [\cfrac { { (x-1) }^{ 3 } }{ 3 } ] }_{ 0 }^{ 1 }$$
$$A=2[0-\cfrac{-1}{3}]$$
$$A=\cfrac{2}{3}\ sq\ units$$
Question 10
1 / -0

Find the area of the region bounded by the curves $${y}^{2}=4ax$$ and $${x}^{2}=4ay$$.

A
$$\dfrac{16}{3}a^2\ sq.unit$$

B
$$\dfrac{8}{3}a^2\ sq. unit$$

C
$$\dfrac{6}{3}a^2\ sq. unit$$

D
None of these

Solution

$${y^2} = 4ax - \left( 1 \right)$$
$${x^2} = 4ay - \left( 2 \right)$$
Solving equation 1 & 2
$${y^2} = 4ax$$
$$ \Rightarrow {\left( {\dfrac{{{x^2}}}{{4a}}} \right)^2} = 4ax$$
$$ \Rightarrow \dfrac{{{x^4}}}{{16{a^2}}} = 4ax$$
$$ \Rightarrow {x^4} = 64{a^3}x$$
$$ \Rightarrow {x^4} - 64{a^3}x = 0$$
$$ \Rightarrow x\left( {{x^3} - 64{a^3}} \right) = 0$$
$$ \Rightarrow x = 0\,\,{x^3} = 64{a^3}\,\,\,x = 4a$$
Area bounded
$$ = \int\limits_0^{4a} {\left( {\sqrt {4ax} - \frac{{{x^2}}}{{4a}}} \right)} dx$$
$$ = \int\limits_0^{4a} {\left( {2\sqrt a \sqrt x - \frac{{{x^2}}}{{4a}}} \right)} dx$$
$$ = 2\sqrt a \int\limits_0^{4a} {\sqrt x dx - \frac{1}{{4a}}\int\limits_0^{4a} {{x^2}dx} } $$
$$ = {\left[ {2\sqrt a \times \frac{{2{x^{3/2}}}}{3}} \right]_0}^{4a} - \frac{1}{{4a}}{\left[ {\frac{{{x^3}}}{3}} \right]_0}^{4a}$$
$$ = \dfrac{{4\sqrt a }}{3}\left[ {{{\left( {4a} \right)}^{3/2}}} \right] - \frac{1}{{12a}}\left[ {64{a^3}} \right]$$
$$ = \dfrac{{4\sqrt a }}{3} \times 8a\sqrt a - \frac{{16}}{{3a}} \times {a^3}$$
$$ = \dfrac{{32{a^2}}}{3} - \frac{{16{a^2}}}{3}$$
$$ = \dfrac{{16}}{3}{a^2}$$