Application of Integrals Test

Result

Application of Integrals Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

If the area of the region bounded by the curves, $$y=x^{2}, y=\dfrac{1}{x}$$ and the lines $$y=0$$ and $$x=t(t>1)$$ is $$1$$ sq. units, then $$t$$ is equal to :

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

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

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

D
$$e^{\dfrac{2}{3}}$$

Solution
Question 2
1 / -0

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

A
$$\frac{{13}}{3}$$

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

C
$$\frac{{10}}{3}$$

D
$$\frac{{5}}{3}$$

Solution
Question 3
1 / -0

The area of the region bounded by the curves $$y = ex \log x$$ and $$y = \dfrac { \log x } { ex }$$ is

A
$$\dfrac { e } { 4 } - \dfrac { 5 } { 4 e }$$

B
$$\dfrac { e } { 4 } + \dfrac { 5 } { 4 e }$$

C
$$\dfrac { e } { 3 } - \dfrac { 5 } { 4 e }$$

D
$$5e$$

Solution
Question 4
1 / -0

The area (in square units) of the region described by $$A={(x,y):y\ge x^{2}-5x+4,x+y\ge 1,y\le 0}$$ is

A
$$\dfrac {19}{6}$$

B
$$\dfrac {17}{6}$$

C
$$\dfrac {7}{2}$$

D
$$\dfrac {13}{6}$$

Solution
Question 5
1 / -0

The area between the curve $$y = 4 + 3 x - x ^ { 2 }$$ and $$x -$$axis is

A
$$125/ 6$$

B
$$125/ 3$$

C
$$125/ $$

D
None
Question 6
1 / -0

Find area of region represented by $$3x+4y > 12, 4x+3y > 12$$ and $$x+y < 4$$.

A
$$2-\dfrac{6}{7}=\dfrac{8}{7}$$

B
$$2+\dfrac{6}{7}=\dfrac{8}{7}$$

C
$$2+\dfrac{6}{7}=\dfrac{7}{8}$$

D
$$\dfrac{6}{7}=\dfrac{7}{8}$$

Solution
Question 7
1 / -0

The area bounded by the hyperbola $$x^2 - y^2 = 4$$ between the lines $$x = 2$$ and $$x = 4$$ is

A
$$4\sqrt{3} - 2 \,log(2 + \sqrt{3})$$

B
$$8\sqrt{3} - 4 \,log(2 - \sqrt{3})$$

C
$$8\sqrt{3} - 4 \,log(2 + \sqrt{3})$$

D
$$4\sqrt{3} - 2 \,log(2 - \sqrt{3})$$

Solution
Question 8
1 / -0

Area of the contained between the parabola $$x^2=4y$$ and the curve $$y=\dfrac{8}{x^2+4}$$ is $$2\pi-K$$ then K=

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

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

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

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

Solution
Question 9
1 / -0

The area enclosed by the curves $$y-\sin x+\cos x$$ and $$y-\left| \cos x-\sin x \right| $$ over the interval $$\{ 0 , \pi / 2 \}$$ is given as $$2\sqrt { a } ( \sqrt { b } - c )$$ where $$a$$ and $$b$$ are prime number then the value of $$a,b$$ and $$c$$ respectively.

A
$$( 2,2,1 )$$

B
$$( 2,2,-1 )$$

C
$$( 3,2,-1 )$$

D
none of these

Solution
Question 10
1 / -0

The area (in sq. units) of the region bounded by the curve, $$12 y = 36 - x ^ { 2 }$$ and the tangents drawn to it at the points,where the curve intersects the $$x$$-axis, is :

A
$$18$$

B
$$27$$

C
$$6$$

D
$$12$$

Solution