Integrals Test

Result

Integrals Test - 40

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

If $$\int _{ log2 }^{ x }{ \dfrac { dx }{ \sqrt { { e }^{ x }-1 } } } =\dfrac { \pi }{ 6 } ,$$then x is equal to _________.

A
4

B
in 8

C
in 4

D
None of these

Solution
Question 2
1 / -0

If $$\int { x\sin { x } } dx=-x\cos { x } +\alpha $$, then $$\alpha$$ is equal to

A
$$\sin x+C$$

B
$$\cos x+C$$

C
$$C$$

D
None of these

Solution

$$I=\displaystyle\int{x\sin{x}dx}$$

Integrating by parts,

Let $$u=x\Rightarrow\,du=dx$$

$$dv=\sin{x}dx\Rightarrow\,v=-\cos{x}$$

$$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ] $$......by parts formula

$$I=-x\cos{x}+\displaystyle\int{\cos{x}dx}$$

$$I=-x\cos{x}+\sin{x}+c$$ ......where $$c$$ is the constant of integration.

We have $$I=\displaystyle\int{x\sin{x}dx}=-x\cos{x}+\alpha$$

Comparing with $$I=-x\cos{x}+\sin{x}+c$$ we have

$$\alpha=\sin{x}+c$$
Question 3
1 / -0

What is $$\displaystyle \int \dfrac{dx}{2x^2 - 2x + 1}$$ equal to ?

A
$$\dfrac{\tan^{-1} (2x - 1)}{2} + c$$

B
$$2 \tan^{-1} (2x - 1) += c$$

C
$$\dfrac{\tan^{-1} (2x + 1)}{2} + c$$

D
$$\tan^{-1} (2x - 1) + c$$

Solution

Given,

$$\int \dfrac{1}{2x^2-2x+1}dx$$

complete the square

$$=\int \dfrac{1}{2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}}dx$$

apply $$u=x-\frac{1}{2}$$

$$\int \dfrac{2}{4u^2+1}du$$

apply $$u=\frac{1}{2}v$$

$$=2\cdot \int \dfrac{1}{2\left(v^2+1\right)}dv$$

$$=2\cdot \dfrac{1}{2}\tan ^{-1} \left(v\right)$$

$$=\tan ^{-1} \left(2\left(x-\dfrac{1}{2}\right)\right)$$

$$=\tan ^{-1} \left(2x-1\right)$$

$$=\tan ^{-1}\left(2x-1\right)+C$$
Question 4
1 / -0

Select and write the most appropriate answer from the given alternatives for question :
If $$\displaystyle \int^k_0 4x^3dx=16$$, then the value of $$k$$ is _____.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Given:
$$\displaystyle \int^K_0 4x^3dx=16$$

$$4\left(\dfrac{x^4}{4}\right)^K_0=16$$

$$(x^4)^k_0=16$$

$$k^4=16$$

$$k^4=2^4$$

$$\boxed{k=2}$$
Question 5
1 / -0

Evaluate $$\displaystyle\int^4_1x\sqrt{x}dx$$

A
$$12.8$$

B
$$12.4$$

C
$$7$$

D
None of these

Solution

$$I=\displaystyle\int^4_1x^{3/2}dx$$

$$=\left[\dfrac{2}{3}x^{5/2}\right]^4_1$$ ............. $$\because\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+c$$

$$=\dfrac{62}{5}=12.4$$.
Question 6
1 / -0

Evaluate : $$\displaystyle\int^1_0\dfrac{dx}{\sqrt{5x+3}}$$

A
$$\dfrac{2}{5}(\sqrt{8}-\sqrt{3})$$

B
$$\dfrac{2}{5}(\sqrt{8}+\sqrt{3})$$

C
$$\dfrac{2}{5}\sqrt{8}$$

D
None of these

Solution

$$I=\displaystyle\int^1_0(5x+3)^{-1/2}dx=\left[2\cdot \dfrac{(5x+3)^{1/2}}{5}\right]^1_0=\dfrac{2}{5}(\sqrt{8}-\sqrt{3})$$.
Question 7
1 / -0

Evaluate : $$\displaystyle\int^2_0\sqrt{6x+4}dx$$

A
$$\dfrac{64}{9}$$

B
$$7$$

C
$$\dfrac{56}{9}$$

D
$$\dfrac{60}{9}$$

Solution
Question 8
1 / -0

Evaluate $$\displaystyle\int^2_0\dfrac{dx}{\sqrt{4-x^2}}$$

A
$$1$$

B
$$\sin^{-1}\dfrac{1}{2}$$

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

D
None of these

Solution
Question 9
1 / -0

Evaluate $$\displaystyle\int^1_0\dfrac{x^3}{(1+x^8)}dx$$

A
$$\dfrac{\pi}{2}$$

B
$$\dfrac{\pi}{4}$$

C
$$\dfrac{\pi}{8}$$

D
$$\dfrac{\pi}{16}$$

Solution

Let $$I=\displaystyle\int^1_0\dfrac{x^3}{(1+x^8)}dx$$

Put $$x^4=t$$ and $$4x^3dx=dt$$.

$$[x=0\Rightarrow t=0]$$ and $$[x=1\Rightarrow t=1]$$

$$\therefore I=\dfrac{1}{4}\displaystyle\int^1_0\dfrac{dt}{(1+t^2)}\\$$
$$=\dfrac{1}{4}[\tan^{-1}t]^1_0$$ ........... $$\because\displaystyle \int \dfrac {dx}{ {x^2+a^2} }=\tan^{-1}\dfrac xa+C\\$$
$$=\dfrac{1}{4}[\tan^{-1}1-\tan^{-1}0]\\$$
$$=\left(\dfrac{1}{4}\times \dfrac{\pi}{4}\right)=\dfrac{\pi}{16}$$.
Question 10
1 / -0

Evaluate $$\displaystyle\int^{\sqrt{8}}_{\sqrt{3}}x\sqrt{1+x^2}dx$$

A
$$\dfrac{19}{3}$$

B
$$\dfrac{19}{6}$$

C
$$\dfrac{38}{3}$$

D
$$\dfrac{9}{4}$$

Solution