Definite Integrals MCQ Quiz - Objective Question with Answer for Definite Integrals - Download Free PDF

Calculation:

Given,

The function is $f(x)=\lfloor x^2\rfloor$.

We are tasked with finding the value of ${\int }_{\sqrt{2}}^{2}f(x)dx$.

We can break the integral into two parts as follows:

${\int }_{\sqrt{2}}^{2}f(x)dx={\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx$

For the range $\sqrt{2}\le x\le \sqrt{3}$, $\lfloor x^2\rfloor =2$. So the first part of the integral is:

${\int }_{\sqrt{2}}^{\sqrt{3}}2dx=2\times (\sqrt{3}-\sqrt{2})$

For the range $\sqrt{3}\le x\le 2$, $\lfloor x^2\rfloor =3$. So the second part of the integral is:

${\int }_{\sqrt{3}}^{2}3dx=3\times (2-\sqrt{3})$

Now, we calculate the values:

$2\times (\sqrt{3}-\sqrt{2})=2\sqrt{3}-2\sqrt{2}$

$3\times (2-\sqrt{3})=6-3\sqrt{3}$

Combining the two parts:

$2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}=6-2\sqrt{2}-\sqrt{3}$

Hence, the correct answer is Option 1.

Calculation:

Given,

The function is $f(x)=\lfloor x^2\rfloor$, where $\lfloor x^2\rfloor$ is the greatest integer function.

We are tasked with finding:

${\int }_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\lfloor x^2\rfloor dx$

Decomposing the integral based on the function:

For \$\frac{\sqrt{3}}{2}\le x<1$, x² lies between $\frac{3}{4}$ and 1, so $\lfloor x^2\rfloor =0$. Therefore,

${\int }_{\frac{\sqrt{3}}{2}}^{1}0dx=0$

For$1\le x<\sqrt{2}$, x²  lies between 1 and 2, so $\lfloor x^2\rfloor =1.$ Therefore,

${\int }_1^{\sqrt{2}}1dx=\sqrt{2}-1$

For $\sqrt{2}\le x<\sqrt{3}$, x²  lies between 2 and 3, so $\lfloor x^2\rfloor =2$ . Therefore,

${\int }_{\sqrt{2}}^{\sqrt{3}}2dx=2(\sqrt{3}-\sqrt{2})$

Summing up all the results:

${\int }_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\lfloor x^2\rfloor dx=0+(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})$

= $2(\sqrt{3}-\sqrt{2})$.

Hence, the correct answer is Option 2. 

Calculation:

The given function is:

$f(x)={\int }_0^2{e}^{|x-t|}dt$

We handle the absolute value function |x - t|  by splitting the integral into two cases based on the value of x  relative to t.

If $x\ge t$, then $|x-t|=x-t$, and if x < t , then |x - t| = t - x .

The function can then be written as:

$f(x)={\int }_0^x{e}^{x-t}dt+{\int }_x^2{e}^{t-x}dt$

First, compute the first integral:

${\int }_0^x{e}^{x-t}dt=e^x(1-e^{-x})$

Now, compute the second integral:

${\int }_x^2{e}^{t-x}dt=e^{-x}(e^2-e^x)$

The total function is:

$f(x)=e^x(1-e^{-x})+e^{-x}(e^2-e^x)$

Thus, simplifying:

$f(x)=e^x+e^{-x}{e}^2-2$

To minimize the function, we differentiate f(x) and set it equal to zero.

$f^{\prime }(x)=e^x-e^{-x}{e}^2$

Set f'(x) = 0:

$e^x=e^{-x}{e}^2$

$e^{2x}=e^2$

Taking the natural logarithm:

$2x=2$ ⇒ x = 1.

Substitute x = 1 into f(x) to find the minimum value:

$f(1)=e+e-2=2e-2$ = 2 (e - 1)

Hence, the correct answer is Option 1

Concept:

$\int \frac{1}{1+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\mathrm{x}$

Calculation:

${\displaystyle {\int }_0^1\frac{1}{1+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=[\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\mathrm{x}{]}_0^1=[\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}1-\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}0]=\frac{\pi }{4}}$

Concept:

Integral properties: Consider a function f(x) defined on x.

• $\underset{-\mathrm{a}}{\overset{\mathrm{a}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=\{\begin{array}{c}2\underset{0}{\overset{\mathrm{a}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x},f\left(\mathrm{x}\right)=f\left(-x\right)\\ 0,f\left(\mathrm{x}\right)=-f\left(-x\right)\end{array}$

Calculation:

Let f(x) = sin x – tan x

Checking the function is odd or even,

f(-x) = sin (-x) – tan (-x)

⇒ f(-x) = – sin x + tan x

⇒ f(-x) = –{sin x – tan x}

⇒ f(-x) = –f(x)

Hence, the function is odd.

And we know that, $\underset{-\mathrm{a}}{\overset{\mathrm{a}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=0$  if f(x) is odd.

∴ ${\int }_{-\pi /4}^{\pi /4}\left(\sin \mathrm{x}-\tan \mathrm{x}\right)\mathrm{d}\mathrm{x}=0$

Concept:

Definite Integral properties:

$\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{a}+\mathrm{b}-\mathrm{x}\right)\mathrm{d}\mathrm{x}$
Calculation:

Let f(x) = x(1 – x)⁹

Now using property, $\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{a}+\mathrm{b}-\mathrm{x}\right)\mathrm{d}\mathrm{x}$

${\int }_0^1\mathrm{x}(1-\mathrm{x}{)}^9\mathrm{d}\mathrm{x}=\underset{0}{\overset{1}{\int }}\left(1-\mathrm{x}\right){\left\{1-\left(1-\mathrm{x}\right)\right\}}^9\mathrm{d}\mathrm{x}$

$\Rightarrow \underset{0}{\overset{1}{\int }}\left(1-\mathrm{x}\right){\mathrm{x}}^9\mathrm{d}\mathrm{x}$

$\Rightarrow \underset{0}{\overset{1}{\int }}\left({\mathrm{x}}^9-{\mathrm{x}}^{10}\right)\mathrm{d}\mathrm{x}$

$\Rightarrow {\left[\frac{{\mathrm{x}}^{10}}{10}-\frac{{\mathrm{x}}^{11}}{11}\right]}_0^1$

⇒ 1/10 – 1/11

⇒ 1/110

∴ The value of integral ${\int }_0^1\mathrm{x}(1-\mathrm{x}{)}^9\mathrm{d}\mathrm{x}$ is 1/110.

Concept:

$\int \frac{\mathrm{d}\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{a}}^2}=\frac{1}{\mathrm{a}}{\tan }^{-1}(\frac{\mathrm{x}}{\mathrm{a}})+\mathrm{c}$

Calculation:

Let I = ${\displaystyle {\int }_0^2{\displaystyle \frac{\mathrm{d}\mathrm{x}}{{\mathrm{x}}^2+4}}}$

= ${\displaystyle {\int }_0^2{\displaystyle \frac{\mathrm{d}\mathrm{x}}{{\mathrm{x}}^2+2^2}}}$

$=[\frac{1}{2}{\tan }^{-1}(\frac{\mathrm{x}}{2}){]}_0^2$

$=[\frac{1}{2}{\tan }^{-1}1-\frac{1}{2}{\tan }^{-1}0]$

$=\frac{1}{2}\times \frac{\pi }{4}-0$

= ${\displaystyle \frac{\pi }{8}}$

Concept:

${\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{a}+\mathrm{b}-\mathrm{x}\right)\mathrm{d}\mathrm{x}$

Calculation: 

Let I = ${\int }_0^{2\pi }\frac{\sin 2\mathrm{x}}{\mathrm{a}-\mathrm{b}\cos \mathrm{x}}\mathrm{d}\mathrm{x}$         ----(1)

Using property f(a + b – x),

I = ${\int }_0^{2\pi }\frac{\sin 2(2\pi -\mathrm{x})}{\mathrm{a}-\mathrm{b}\cos (2\pi -\mathrm{x})}\mathrm{d}\mathrm{x}$   

As we know,  sin (2π - x) = - sin x and cos (2π - x) = cos x

I = ${\int }_0^{2\pi }\frac{-\sin 2\mathrm{x}}{\mathrm{a}-\mathrm{b}\cos \mathrm{x}}\mathrm{d}\mathrm{x}$         ----(2)       

I = -I

2I = 0

∴ I = 0

Concept:

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Calculation: 

I = ${\int }_1^{\mathrm{\infty }}\frac{4}{{\mathrm{x}}^4}\mathrm{d}\mathrm{x}$

= ${\int }_1^{\mathrm{\infty }}4{\mathrm{x}}^{-4}\mathrm{d}\mathrm{x}$

= ${\left[\frac{4{\mathrm{x}}^{-3}}{-3}\right]}_1^{\mathrm{\infty }}$

= $\frac{-4}{3}{\left[\frac{1}{{\mathrm{x}}^3}\right]}_1^{\mathrm{\infty }}$

= $\frac{-4}{3}[\frac{1}{\mathrm{\infty }}-\frac{1}{1}]$

= $\frac{-4}{3}[0-1]$

= $\frac{4}{3}$

Concept:

${\displaystyle {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}={\displaystyle {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})\mathrm{d}\mathrm{x}}}$

Calculations:

Consider, I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{\sin \mathrm{x}}}{\sqrt{\sin \mathrm{x}}+\sqrt{\cos \mathrm{x}}}}\mathrm{d}\mathrm{x}}$             ....(1)

I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{\sin ({\displaystyle \frac{\pi }{2}}-\mathrm{x})}}{\sqrt{\sin ({\displaystyle \frac{\pi }{2}}-\mathrm{x})}+\sqrt{\cos ({\displaystyle \frac{\pi }{2}}-\mathrm{x})}}}\mathrm{d}\mathrm{x}}$

I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{\cos \mathrm{x}}}{\sqrt{\cos \mathrm{x}}+\sqrt{\sin \mathrm{x}}}}\mathrm{d}\mathrm{x}}$                           ....(2)

Adding (1) and (2), we have

2I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{\cos \mathrm{x}}+\sqrt{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}{\sqrt{\cos \mathrm{x}}+\sqrt{\sin \mathrm{x}}}}\mathrm{d}\mathrm{x}}$

2I = ${\displaystyle {\int }_0^{\pi /2}\mathrm{d}\mathrm{x}}$

2I = $[\mathrm{x}{]}_0^{\frac{\pi }{2}}$

I = ${\displaystyle \frac{\pi }{4}}$

Concept:

${\displaystyle \frac{\mathrm{d}(\ln \mathrm{x})}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{1}{\mathrm{x}}}$

Calculation:

Let I = ${\int }_1^{\mathrm{e}}\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}\sqrt{2+\ln \mathrm{x}}}$

Let (2 + ln x) = t²

Differentiating with respect to x, we get

⇒ (0 + $\frac{1}{\mathrm{x}}$)dx = 2tdt

⇒ $\frac{1}{\mathrm{x}}$dx = 2tdt

Now,

$$\begin{gathered} \mathrm{I}={\int }_{\sqrt{2}}^{\sqrt{3}}\frac{2\mathrm{t}\mathrm{d}\mathrm{t}}{\sqrt{{\mathrm{t}}^2}} \\ =2{\int }_{\sqrt{2}}^{\sqrt{3}}\frac{\mathrm{t}\mathrm{d}\mathrm{t}}{\mathrm{t}} \\ =2{\int }_{\sqrt{2}}^{\sqrt{3}}\mathrm{d}\mathrm{t} \\ =2[\mathrm{t}{]}_{\sqrt{2}}^{\sqrt{3}} \\ =2(\sqrt{3}-\sqrt{2}) \end{gathered}$$

Concept:

${\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{a}+\mathrm{b}-\mathrm{x}\right)\mathrm{d}\mathrm{x}$

Calculation: 

Let I = ${\int }_0^{\pi }{\sin }^6\mathrm{x}{\cos }^5\mathrm{x}\mathrm{d}\mathrm{x}$             .... (1)

Using property f(a + b – x),

I = ${\int }_0^{\pi }{\sin }^6(\pi -\mathrm{x}){\cos }^5(\pi -\mathrm{x})\mathrm{d}\mathrm{x}$

As we know,  sin (π - x) = sin x and cos (π - x) = -cos x

I = -${\int }_0^{\pi }{\sin }^6\mathrm{x}{\cos }^5\mathrm{x}\mathrm{d}\mathrm{x}$                .... (2)

I = -I

2I = 0

∴ I = 0

Concept:

f(x) = |x| will be equal to 

• x, if x > 0
• -x, if x < 0
• 0, if x = 0

∫ dx = x + C  (C is a constant)

∫ xⁿ dx = xⁿ⁺¹/n+1 + C

Calculation:

Let, $I={\int }_{-2}^{-1}\frac{x}{|x|}dx$

Using the above concept, as x ∈ (-2, -1)

⇒ $I={\int }_{-2}^{-1}\frac{x}{-x}dx$

⇒ $I=-1{\int }_{-2}^{-1}(1)dx$

⇒ $I=-[x{]}_{-2}^{-1}$  

⇒ I = -[-1 - (-2)] 

∴  ${\int }_{-2}^{-1}\frac{x}{|x|}dx$ = -1

Concept:

Integral property:

• ∫ xn dx = $\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}$+ C ; n ≠ -1
• $\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\ln \mathrm{x}$ + C
• ∫ ex dx = ex+ C

Calculation:

I = $\int \frac{2\mathrm{x}+1}{{\left({\mathrm{x}}^2+\mathrm{x}+1\right)}^2}\mathrm{d}\mathrm{x}$

Let x2 + x + 1 = t

⇒ (2x + 1) dx = dt

I = $\int \frac{\mathrm{d}\mathrm{t}}{{\mathrm{t}}^2}$

I = $\frac{{\mathrm{t}}^{-1}}{-1}$

I = $\frac{-1}{\mathrm{t}}$

I = $\frac{-1}{{\mathrm{x}}^2+\mathrm{x}+1}$

Putting limits

I = ${\left[\frac{-1}{{\mathrm{x}}^2+\mathrm{x}+1}\right]}_{-1}^1$

I = $\frac{-1}{1^2+1+1}-\left(\frac{-1}{(-1{)}^2+(-1)+1}\right)$

I = $1-\frac{1}{3}$ = $\text{boldsymbol}\frac{2}{3}$

Concept:

${\int }_{\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=0$

Calculation:

We know, 

${\int }_{\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=0$

Here, limit of integration is same (i.e., π/4)

$\therefore {\int }_{\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\cos \left({\mathrm{e}}^{3\mathrm{x}}\right)}{{\mathrm{x}}^4+{\mathrm{x}}^3+1}=0$

Hence, option (3) is correct.