Application of Integrals Test

Result

Application of Integrals Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The area (in sq. units) in the first quadrant bounded by the parabola, $$y = x^2 + 1$$, the tangent to it at the point $$(2, 5)$$ and the coordinate axes is:-

A
$$\dfrac{14}{3}$$

B
$$\dfrac{187}{24}$$

C
$$\dfrac{37}{24}$$

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

Solution
Question 2
1 / -0

The area of the region bounded by $$y=\left | x-1 \right | and \,\,y=1 $$ is

A
1

B
2

C
1/2

D
None of these

Solution
Question 3
1 / -0

The area of the region
$$A=[(x,y):0\le y\le x|x|+1$$ and $$-1\le x\le 1]$$ in sq . units is :

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

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

C
$$2$$

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

Solution

The graph is as follows:
$$\displaystyle \int_{-1}^0(-x^2+1)dx+\displaystyle \int_0^1(x^2+1)dx=2$$
Question 4
1 / -0

The slope of the tangent to the curve y =f(x) at a point (x, Y) is 2x + 1 and the curve passes through (1, 2) The area of the region bounded by the curve, the x-axis and the line x= 1 is -

A
5/3 units

B
5/6 units

C
6/5 units

D
6 units

Solution
Question 5
1 / -0

The area (in sq. units) of the region bounded by the parabola, $$y = x ^ { 2 } + 2$$ and the lines, $$y = x + 1 , x = 0$$ and $$x = 3 ,$$ is :

A
$$\dfrac { 15 } { 4 }$$

B
$$\dfrac { 15 } { 2 }$$

C
$$\dfrac { 21 } { 2 }$$

D
$$\dfrac { 17 } { 4 }$$

Solution

Req. area $$= \int _ { 0 } ^ { 3 } \left( x ^ { 2 } + 2 \right) d x - \dfrac { 1 } { 2 } \cdot 5.3 = 9 + 6 - \dfrac { 15 } { 2 } = \dfrac { 15 } { 2 }$$
Question 6
1 / -0

The area (in sq. units) of the region $$\{ { x },{ y }):{ y }^{ { 2 } }\geq 2{ x }$$ and $$x ^ { 2 } + y ^ { 2 } \leq 4 x , x \geq 0 , y \geq 0 \}$$ is :

A
$$\pi - \dfrac { 4 \sqrt { 2 } } { 3 }$$

B
$$\dfrac { \pi } { 2 } - \dfrac { 2 \sqrt { 2 } } { 3 }$$

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

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

Solution
Question 7
1 / -0

The area of the region $$\left\{ ( x , y ) : x ^ { 2 } + y ^ { 2 } \leq 1 \leq x + y \right\}$$ is

A
$$\dfrac { \pi ^ { 2 } } { 5 } \text { unit } ^ { 2 }$$

B
$$\dfrac { \pi ^ { 2 } } { 2 } \text { unit } ^ { 2 }$$

C
$$\dfrac { \pi ^ { 2 } } { 3 } \text { unit } ^ { 2 }$$

D
$$\left( \dfrac { \pi } { 4 } - \dfrac { 1 } { 2 } \right)\text { unit } ^ { 2 }$$

Solution
Question 8
1 / -0

The area of the region bounded by the parabola y = $$x^2$$ 3x with y 0 is

A
$$3$$

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

C
$$\dfrac{9}{2}$$

D
$${9}{}$$

Solution
Question 9
1 / -0

The area of the quadrilateral formed by the tangents at the endpoints of the latus recta to the ellipse, $$\dfrac{x^{2}}{9}+\dfrac{y^{2}}{5}=1$$ is

A
$$\dfrac{27}{4}$$

B
$$18$$

C
$$\dfrac{27}{2}$$

D
$$27$$

Solution

Given equation of ellipse is
$$\dfrac{{x}^{2}}{9}+\dfrac{{y}^{2}}{5}=1$$ ......$$(1)$$
$$\therefore {a}^{2}=9,{b}^{2}=5$$
$$\Rightarrow a=3,b=\sqrt{5}$$
Now, $$e=\sqrt{1-\dfrac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\dfrac{5}{9}}=\dfrac{2}{3}$$
Foci$$=\left(\pm ae,0\right)=\left(\pm 2,0\right)$$
and $$\dfrac{{b}^{2}}{a}=\dfrac{5}{3}$$
$$\therefore$$ Extremities of one of latus rectum are $$\left(2,\dfrac{5}{3}\right)$$ and $$\left(2,\dfrac{-5}{3}\right)$$
$$\therefore$$ Equation of tangent at $$\left(2,\dfrac{5}{3}\right)$$ is
$$\dfrac{x\left(2\right)}{9}+\dfrac{y\left(\dfrac{5}{3}\right)}{5}=1$$
or $$2x+3y=9$$ .......$$(2)$$
Eqn$$(2)$$ intersects $$X$$ and $$Y$$ axes at $$\left(\dfrac{9}{2},0\right)$$ and $$\left(0,3\right)$$ respectively.
$$\therefore$$ Area of quadrilateral $$4\times$$Area of $$\triangle{POQ}$$
$$=4\times\dfrac{1}{2}\times\dfrac{9}{2}\times 3=27$$sq.units.
Question 10
1 / -0

The area bounded by the curves $$x+2|y|=1$$ and $$x=0$$ is?

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{2}$$

C
$$1$$

D
$$2$$

Solution

$$x+2\left| y \right| =1$$ and $$x=0$$
$$y=+ve$$ $$x+2y=1$$
$$y=-ve$$ $$x-2y=1$$
Area of triangle $$=\cfrac{1}{2}\times base\times height\\=\cfrac{1}{2}\times 1\times 1\\=\cfrac{1}{2}\text{ }unit$$