Application of Integrals Test

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Application of Integrals Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

The area enclosed between the curves $$y=log_e(x+e), x=log_e\left(\dfrac{1}{y}\right)$$ and the x-axis is?

A
$$2e$$

B
$$e$$

C
$$4e$$

D
None of these

Solution
Question 2
1 / -0

The area (in sq. units) bounded by the parabola $$y=x^{2}-1$$ , the tangent at the point (2,3) to it and the y-axis is:

A
$$\frac{14}{3}$$

B
$$\frac{56}{3}$$

C
$$\frac{8}{3}$$

D
$$\frac{32}{3}$$

Solution

Equation of tangent at $$(2,3)$$ on
$$y=x^2-1$$, is $$y=(4x-5).......(i)$$
$$\therefore$$ Required shaded area
$$= ar(\triangle ABC)- \int\limits_{-1}^3 \sqrt{y+1}dy$$
$$= \dfrac12. (8).(2)- \dfrac23 ((y+1)^{3/2})_{-1}^3$$
$$= 8- \dfrac{16}{3}= \dfrac83$$
Question 3
1 / -0

The area bounded by the curve $$y \le x^2 + 3x , 0 \le y \le 4, \, 0 \le x \le 3$$ , is

A
$$\dfrac{59}{6}$$

B
$$\dfrac{57}{4}$$

C
$$\dfrac{59}{3}$$

D
$$\dfrac{57}{6}$$

Solution

$$y = x^2 + 3x = 4 \Rightarrow x^2 + 3x - 4 = 0$$
$$(x + 4) (x - 1) = 0$$
$$x = 1$$
area $$= \displaystyle \int_0^1 (x^2 + 3x ) dx + 2(4) = \dfrac{1}{3} + \dfrac{3}{2} + 8 = \dfrac{11}{6} + 8 = \dfrac{59}{6}$$
Question 4
1 / -0

The area bounded by the curve $$y=x$$ $$X-$$axis and the lines $$x=-1$$ and $$x=1$$ is

A
$$0$$

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

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

D
$$\dfrac {4}{3}$$
Question 5
1 / -0

The area of the region bounded by the curve $$y=\phi(x),y=0$$ and $$x=10$$ is

A
$$\dfrac {81}{4}$$

B
$$\dfrac {79}{4}$$

C
$$\dfrac {73}{4}$$

D
$$19$$
Question 6
1 / -0

If the area (in sq. units) of the region $$\{(x, y): y^2\le 4x, x+y\le 1, x\geq 0, y\ge 0 \}$$ is $$a\sqrt{2}+b$$, then $$a-b$$ is equal to?

A
$$\dfrac{8}{3}$$

B
$$\dfrac{10}{3}$$

C
$$6$$

D
$$-\dfrac{2}{3}$$

Solution

$$\{(x, y): y^2\leq 4x, x+y\leq 1, x\geq 0, y\geq 0\}$$
$$A\displaystyle\int^{3-2\sqrt{2}}_02\sqrt{x}dx+\dfrac{1}{2}(1-(3-2\sqrt{2}))(1-(3-2\sqrt{2}))$$
$$=\dfrac{2[x^{3/2}]_0^{3-2\sqrt{2}}}{3/2}+\dfrac{1}{2}(2\sqrt{2}-2)(2\sqrt{2}-2)$$
$$=\dfrac{8\sqrt{2}}{3}+\left(-\dfrac{10}{3}\right)$$
$$a=\dfrac{8}{3}$$, $$b=-\dfrac{10}{3}$$
$$a-b=6$$.
Question 7
1 / -0

Let $$f(x, y)=\{(x, y): y^2 \leq 4x, 0\leq x\leq \lambda\}$$ and $$s(\lambda)$$ is area such that $$\dfrac{S(\lambda)}{S(4)}=\dfrac{2}{5}$$. Find the value of $$\lambda$$.

A
$$4\left(\dfrac{4}{25}\right)^{1/3}$$

B
$$4\left(\dfrac{2}{25}\right)^{1/3}$$

C
$$2\left(\dfrac{4}{25}\right)^{1/3}$$

D
$$2\left(\dfrac{2}{25}\right)^{1/3}$$

Solution

$$y^2=4x$$
$$S(I)=\left.2\displaystyle\int^{\lambda}_{0}2\sqrt{x}dx=\dfrac{4x^{3/2}}{3/2}\right]^{\lambda}_{0}=\dfrac{8}{3}\lambda^{3/2}$$
$$\dfrac{S(\lambda)}{S(4)}=\dfrac{2}{5}$$ $$\Rightarrow \dfrac{\lambda^{3/2}}{4^{3/2}}=\dfrac{2}{5}$$
$$\lambda =4\left(\dfrac{2}{5}\right)^{2/3}=4\left(\dfrac{4}{25}\right)^{1/3}$$.
Question 8
1 / -0

Region formed by $$|x-y| \le 2$$ and $$|x + y| \le 2$$ is

A
Rhombus of side is $$2$$

B
Square of area is $$6$$

C
Rhombus of area is $$8\sqrt{2}$$

D
Square of side is $$2\sqrt{2}$$

Solution

$$\dfrac{\text{ABCD is a square}}{Area = 4 \times \dfrac{1}{2}\times 2 \times 2 = 8}$$
side $$= 2 \sqrt{2}$$
Question 9
1 / -0

The region represented by $$|x - y| \le 2$$ and $$|x + y| \le 2$$ is bounded by a:

A
Square of side length $$2\sqrt{2}$$ units

B
Rhombus of side length $$2$$ units

C
Square of area $$16 \,sq$$ units

D
Rhombus of area $$8\sqrt{2} sq.$$ units

Solution

$$|x - y| \le 2$$ and $$|x + y| \le 2$$

$$\Rightarrow x-y\le 2$$ or $$x-y \ge 2$$ and $$x+y \le 2$$ and $$x+y \ge-2$$

After plotting them on the graph the common region is

Square whose side is $$2\sqrt{2}$$
Question 10
1 / -0

The area (in sq. units) of the region $$A=\{(x, y):x^2\leq y\leq x+2\}$$ is?

A
$$\dfrac{10}{3}$$

B
$$\dfrac{9}{2}$$

C
$$\dfrac{31}{6}$$

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

Solution

$$x^2\leq y\leq x+2$$
$$x^2=y; y=x+2$$
$$x^2=x+2$$
$$x^2-x-2=0$$
$$(x-2)(x-1)=0$$
$$x=2, -1$$
Area $$=\displaystyle\int^2_{-1}(x+2)-x^2dx=\left[\dfrac {x^2}{2}+2x-\dfrac {x^3}{3}\right]^2_{-1}$$

$$=2+4-\dfrac 83-\dfrac 12 +2 -\dfrac 13=\dfrac{9}{2}$$.