Application of Integrals Test

Result

Application of Integrals Test - 38

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The area (in sq units) of the region $$\{ (x,y):{ y }^{ 2 }\ge 2x$$ and $${ x }^{ 2 }+{ y }^{ 2 }\le 4x ,\chi \ge 0, Y\ge 0\}$$

A
$$\pi -\frac { 4 }{ 3 } $$

B
$$\pi -\frac { 8 }{ 3 } $$

C
$$\pi -\frac { 4\sqrt { 2 } }{ 3 } $$

D
$$-\frac { 2\sqrt { 2 } }{ 3 } $$

Solution
Question 2
1 / -0

The area bounded by the curve $$xy^{2}=1$$ and the lines $$x=1$$, $$x=2$$ is

A
$$4\left( {\sqrt 2 - 1} \right)$$

B
$$4\left( {\sqrt 2 + 1} \right)$$

C
$$2\left( {\sqrt 2 - 1} \right)$$

D
$$2\left( {\sqrt 2 + 1} \right)$$

Solution
Question 3
1 / -0

The area in square units bounded by the curves $$y = x ^ { 3 } , y = x ^ { 2 }$$ and the ordinates $$x = 1 , x = 2$$ is

A
$$\frac { 17 } { 12 }$$

B
$$\frac { 12 } { 13 }$$

C
$$\frac { 2 } { 7 }$$

D
$$\frac { 7 } { 2 }$$

Solution

$$\int _{ 1 }^{ 2 }{ \left( { x }^{ 3 }{ -x }^{ 2 } \right) dx } \\ ={ \left[ \frac { { x }^{ 4 } }{ 4 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 1 }^{ 2 }\\ =\left[ 4-\frac { 8 }{ 3 } \right] -\left[ \frac { 1 }{ 4 } -\frac { 1 }{ 3 } \right] \\ =\left[ \frac { 4 }{ 3 } \right] -\left[ -\frac { 1 }{ 12 } \right] \\ =\frac { 4 }{ 3 } +\frac { 1 }{ 12 } \\ =\frac { 17 }{ 12 } $$
Question 4
1 / -0

The area bounded by the curves $$y=\sqrt{-x}$$ and $$x=-\sqrt{-y}$$, where $$x,y\le0$$, is equal to

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

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

C
$$\dfrac{1}{2}sq. unit$$

D
$$Cannot\ be\ determined$$

Solution

Given the curve
$$y=-\sqrt{-x}$$ and $$x=-\sqrt{-y}$$
Now, the area bound by the both curve
$$=\displaystyle \int_{-1}^{0}{(-x^2+\sqrt{-x})dx}$$
$$=\displaystyle \int_{-1}^{0}{-x^2 dx}+\displaystyle \int_{-1}^{0}{\sqrt{-x}dx}$$
change
$$=\displaystyle \int_{-1}^{0}{\sqrt{x} dx}-\displaystyle \int_{-1}^{0}{x^2 dx}$$
$$=\left[\dfrac{2}{3}x^{3/2}-\dfrac{x^3}{3}\right]_0^1$$
$$=\left(\dfrac{2}{3}-\dfrac{1}{3}\right)-(0-0)$$
$$=\dfrac{1}{3}\ sq.\ unit$$
Hence, this is the answer.
Question 5
1 / -0

Tangents are drawn from a point $$P$$ to a parabola $$y^{2}=4ax$$. The area enclosed by the tangents and the corresponding chord of contact is $$4a^{2}$$. Then point $$P$$ satisfies

A
$$y^{2}=4ax$$

B
$$y^{2}=2a(x+a)$$

C
$$y^{2}=4a(x-a)$$

D
$$y^{2}=4a(x+a)$$
Question 6
1 / -0

Let $$T$$ be the triangle with vertices $$\left (0,0\right), \left (0,{c}^{2}\right)\ and \left (c,{c}^{2}\right)$$ and let $$R$$ be the region between $$y=cx$$ and $$y={x}^{2}\ where c>0$$ then

A
Area $$\left( R \right) =\dfrac { { c }^{ 3 } }{ 6 }$$

B
Area of $$R=\dfrac {e^{x}}3{3}$$

C
$$\lim_{c \rightarrow 0}\dfrac {Area(T)}{Area(R)}=3$$

D
$$\lim_{c \rightarrow 0}\dfrac {Area(T)}{Area(R)}=\dfrac {3}{2}$$

Solution
Question 7
1 / -0

Let $$f(x)=minimum (x+1,\sqrt{1-x}) $$ for all $$x \le 1$$. Then the area bounded by $$y=f(x)$$ and the x-axis is

A
$$\dfrac{7}{3}$$ sq. units

B
$$\dfrac{1}{6}$$ sq. units

C
$$\dfrac{11}{6}$$ sq. units

D
$$\dfrac{7}{6}$$ sq. units

Solution
Question 8
1 / -0

The area of the figure bounded by the curves $$y ^ { 2 } = 2 x + 1$$ and $$x - y - 1 = 0$$ is

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

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

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

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

Solution
Question 9
1 / -0

Area (in $$sq.$$ unit) of region bounded by $$y=2\cos x,\ y=3\tan x$$ and $$y-$$axis is

A
$$1+3ln \left(\dfrac {2}{\sqrt {3}}\right)$$

B
$$1+\dfrac {3}{2}ln3-3ln2$$

C
$$1+\dfrac {3}{2}ln3-ln2$$

D
$$ln \left(\dfrac {3}{2}\right)$$

Solution

Given that,
$$y=2\cos x, y=3\tan x$$
$$2\cos x=3\tan x$$
$$2\cos x=3\dfrac{\sin x}{\cos x}$$
$$2\cos^2x=3\sin x$$
$$2(1-\sin^2x)=3\sin x$$
$$2-2\sin^2x=3\sin x$$
$$2\sin^2x+3\sin x-2=0$$
On solving,
$$x=\dfrac{\pi}{6}$$ and y-axis
So, $$x=0$$
then,
$$\displaystyle\int^{\pi/6}_0(2\cos x-3\tan x)dx$$
$$=2\displaystyle\int^{\pi/6}_0\cos xdx-3\displaystyle\int^{\pi/6}_0\tan xdx$$
$$=2[\sin x]^{\pi/6}_0-3[log \sec x]^{\pi/6}_0$$
$$=2\left[\sin\dfrac{\pi}{6}-\sin\theta\right]-3\left[log\sec\dfrac{\pi}{6}-log\sec 0\right]$$
$$=2\left[\dfrac{1}{2}-0\right]-3\left[log \dfrac{2}{\sqrt{3}}-log 1\right]$$
$$=\dfrac{2}{2}-0-3\left[log 2-log\sqrt{3}-0\right]$$
$$=1-3[log 2-log \sqrt{3}]$$
$$=1-3log 2+3log \sqrt{3}$$
$$=1-3log 2+3log (3)^{\dfrac{1}{2}}$$
$$=1-3log 2+\dfrac{3}{2}log 3$$
$$=1+\dfrac{3}{2}log 3-3log 2$$.
Question 10
1 / -0

Let $$f\left( x \right)$$ be a non-negative continuous function such that the area bounded by the curve $$y= f\left( x \right) $$, x-axis and the ordinates $$x=\cfrac { \pi }{ 4 } $$, $$x=\beta >\cfrac { \pi }{ 4 } $$ is $$\left( \beta \sin { \beta } +\cfrac { \pi }{ 4 } \cos { \beta } +\sqrt { 2 } \beta -\cfrac { \pi }{ 2 } \right) $$. Then $$f\left( \cfrac { \pi }{ 2 } \right) $$ is

A
$$\left( 1-\dfrac { \pi }{ 4 } -\sqrt { 2 } \right)$$

B
$$\left( 1-\dfrac { \pi }{ 4 } +\sqrt { 2 } \right)$$

C
$$\left( \cfrac { \pi }{ 4 } +\sqrt { 2 } -1 \right)$$

D
$$\left( \cfrac { \pi }{ 4 } -\sqrt { 2 } +1 \right)$$

Solution

Given that,
$$\displaystyle\int^{\beta}_{\pi/4}f(x)dx=\beta \sin\beta +\dfrac{\pi}{4}\cos\beta +\sqrt{2}\beta -\dfrac{\pi}{2}$$
Differentiating w.r.t x and we get
$$f(\beta)=\beta \cos\beta +\sin\beta -\dfrac{\pi}{4}\sin\beta +\sqrt{2}-0$$
$$f\left(\dfrac{\pi}{2}\right)=\left(1-\dfrac{\pi}{4}\right)\sin\dfrac{\pi}{2}+\sqrt{2}$$
$$=\left(1-\dfrac{\pi}{4}\right)\times 1+\sqrt{2}$$
$$=1-\dfrac{\pi}{4}+\sqrt{2}$$
Hence, this is the answer.