Application of Integrals Test

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Application of Integrals Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

The area of the figure formed by $$ |x| + |y| = 2$$ is (in sq. units)

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution
Question 2
1 / -0

The area bounded by the curve $$y=\ln (x)$$ and the lines $$y=\ln (3),y=0$$ and $$x=0$$ is equal

A
$$3$$

B
$$3\ln (3)-2$$

C
$$3\ln (3)+2$$

D
$$2$$

Solution
Question 3
1 / -0

The area (in sq.units) of the region described by $$\left\{(x,y):y^{2}\le 2x\ and\ y\ge 4x-1\right\}$$ is

A
$$\dfrac{15}{64}$$

B
$$\dfrac{9}{32}$$

C
$$\dfrac{7}{32}$$

D
$$\dfrac{5}{64}$$

Solution
Question 4
1 / -0

The area bounded by the curves $$y = \log _ { e } x$$ and $$y = \left( \log _ { e } x \right) ^ { 2 }$$ is

A
$$3 - e$$

B
$$e-3$$

C
$$\frac { 1 } { 2 } ( 3 - e )$$

D
$$\frac { 1 } { 2 } ( e - 3 )$$

Solution

$$\begin{array}{l} y=\left( { { { \log }_{ e } }x } \right) \, \, \, \, \, \, \, ----\left( 1 \right) \\ y={ \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, \, \, \, -----\left( 2 \right) \\ For\, the\, po{ { int } }\, of\, { { int } }er\sec tion \\ { \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, =\left( { { { \log }_{ e } }x } \right) \\ { \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, -\left( { { { \log }_{ e } }x } \right) =0 \\ \left( { { { \log }_{ e } }x } \right) \left[ { \left( { { { \log }_{ e } }x } \right) \, -1 } \right] \, \\ \left( { { { \log }_{ e } }x } \right) =0\, \, and\, \left( { { { \log }_{ e } }x } \right) =1 \\ x=1\, \, and\, \, x=e \\ area\, bounded \\ =\left| { \int _{ 1 }^{ e }{ \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } } \right| \, \, \\ consider \\ I=\int { \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } \\ { \log _{ e } }x=t \\ x={ e^{ t } } \\ dx={ e^{ t.dt } } \\ Now, \\ I=\int { { e^{ t } }\left\{ { { t^{ 2 } }-t } \right\} dt } \\ ={ e^{ t } }\left\{ { { t^{ 2 } }-t } \right\} -\left\{ { \left( { 2t-1 } \right) { e^{ t } }-2{ e^{ t } } } \right\} \, \, \, \, \left[ { by\, { { int } }ergration\, by\, parts } \right] \\ ={ e^{ t } }\left[ { { t^{ 2 } }-t-2t+1+2 } \right] \\ ={ e^{ t } }\left[ { { t^{ 2 } }-3t+3 } \right] \\ =x\left[ { \log { { \left( x \right) }^{ 2 } } -3\log \left( x \right) +3 } \right] \\ Now, \\ \left| { \int _{ 1 }^{ e }{ \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } } \right| \, \, \\ =\left| { \left[ { x\left\{ { loh{ { \left( x \right) }^{ 2 } } } \right\} -3\log \left( x \right) +3 } \right] _{ 1 }^{ e } } \right| \\ =\left| { e\left( { 1-3+3 } \right) -1\left( { 0-0+3 } \right) } \right| \\ =\left| { e-3 } \right| \\ =3-e \end{array}$$
Hence, the option $$B$$ is the correct answer.
Question 5
1 / -0

The area common to the parabola $$y=2{ x }^{ 2 }\quad$$ and $$\quad y={ x }^{ 2 }+4$$

A
$$\dfrac { 2 }{ 3 } sq.\quad units\quad $$

B
$$\dfrac { 3 }{ 2 } sq.\quad units\quad $$

C
$$\dfrac { 32 }{ 3 } sq.\quad units\quad $$

D
none of these.

Solution

Find the P.O.I of the parabolas equate the equations $$y = 2x^2$$ and $$y = x^2 + y$$ we get
$$2x^2 = x^2 + 4$$
$$x^2 = 4$$
$$x = \pm 2$$
$$y = 8$$
$$\therefore $$ POI are $$A(-2, 8) $$ & $$C(2, 8)$$
$$\therefore$$ the required are ABCD
$$A = \displaystyle \int_{-2}^2 (y_1 - y_2) dx$$ (where $$y_1 = x^2 + 4 \, \& \, y_2 = 2x^2$$)
$$= \displaystyle \int_{-2}^2 (x^2 + 4 - 2x^2) dx = \int_{-2}^2 (4 - x^2) dx = \left(4x - \dfrac{x^3}{3} \right)^2_{-2}$$
$$\left[4(2) - \dfrac{(2)^3}{3} \right] - \left[4(-2) - \dfrac{(-2)^3}{3} \right] = \left[8 - \dfrac{8}{3} \right] - \left[8 + \dfrac{8}{3} \right]$$
$$= \dfrac{32}{3} $$ sq. units
Question 6
1 / -0

The area bounded by a the curves y=x(1-/nX) and positive X-axis between $$X={ e }^{ -1 }$$ amd X=e is:-

A
$$\left( \frac { { e }^{ 2 }-{ 4e }^{ -2 } }{ 5 } \right) $$

B
$$\left( \frac { { e }^{ 2 }-{ 5e }^{ -2 } }{ 4 } \right) $$

C
$$\left( \frac { { 4e }^{ 2 }-{ e }^{ -2 } }{ 5 } \right) $$

D
$$\left( \frac { { 5e }^{ 2 }-{ e }^{ -2 } }{ 4 } \right) $$

Solution
Question 7
1 / -0

The area bounded by the curves $$y=x(x-3)^{2}$$ and $$y=x$$ is (in $$sq.units$$) is

A
$$28$$

B
$$32$$

C
$$4$$

D
$$8$$

Solution
Question 8
1 / -0

The area bounded by the curve $$y={ x }^{ 2 }$$, X=axis and the ordinates z=1, z=3 is ____________.

A
$$\dfrac { 26 }{ 3 } sq.units$$

B
$$\dfrac { 28 }{ 3 } sq.unit$$

C
$$\dfrac { 1 }{ 3 } sq.units$$

D
$$9\quad sq.units$$
Question 9
1 / -0

The area bounded by the curves $$y=x(x-3)^{2}$$ and $$y=x$$ is (in sq.units) is

A
$$28$$

B
$$32$$

C
$$4$$

D
$$8$$

Solution
Question 10
1 / -0

The area bounded by the curve y = log x, X-axis and the ordinates x =1, x =2 is

A
log 4 sq. units

B
log 2 sq units

C
(log 4 - 1) sq.units

D
(log 4 + 1)sq. units

Solution