Integral Calculus MCQ Quiz - Objective Question with Answer for Integral Calculus - Download Free PDF

Explanation: 

Integration of Power of Tangent functions : 

$I_n=\int ta{n}^{n}xdx$  

Reduction Formula is 

$I_n=\frac{1}{n-1}{\tan }^{n-1}(x)-I_{n-2}$   

Hence Option(1) is the correct answer.

Explanation:

Using the given reduction formula:

$I_n=-\frac{1}{n}{\sin }^{n-1}(x)\cos (x)+\frac{n-1}{n}{I}_{n-2}$ ,
  
for n = 4 ,  

$I_4=-\frac{1}{4}{\sin }^3(x)\cos (x)+\frac{3}{4}{I}_2$  

Using   ${\sin }^2(x)=\frac{1-\cos (2x)}{2}$
  
So,    $I_2=\int {\sin }^2(x)dx=\int \frac{1-\cos (2x)}{2}dx=\frac{x}{2}-\frac{\sin (2x)}{4}$   

Substitute  $I_2$  into the expression for $I_4$ :  

$I_4=-\frac{1}{4}{\sin }^3(x)\cos (x)+\frac{3}{4}(\frac{x}{2}-\frac{\sin (2x)}{4})$   

Simplify,   $I_4=-\frac{1}{4}{\sin }^3(x)\cos (x)+\frac{3x}{8}-\frac{3}{16}\sin (2x)$

Hence Option(2) is the correct answer.

Explanation:
The integral of f(x) being zero does not imply f(x) is identically zero;

f(x) may take both positive and negative values, so its areas cancel out

For example, f(x) = x - 0.5 satisfies the condition

Hence Option(3) is the correct answer.

Explanation: 

$f(x)=x^x$   as  $f(x)=e^{x\ln (x)}$ :  

$\underset{x\to 0^{+}}{lim}x\ln (x)=\underset{x\to 0^{+}}{lim}\frac{\ln (x)}{1/x}$    
  
Using L 'Hôpital's Rule:  

$\underset{x\to 0^{+}}{lim}\frac{\ln (x)}{1/x}=\underset{x\to 0^{+}}{lim}\frac{1/x}{-1/x^2}=\underset{x\to 0^{+}}{lim}-x=0$   
  
Thus,  $\underset{x\to 0^{+}}{lim}f(x)=e^0=1$  

Hence Option(2) is the correct answer.

Explanation:

S be that part of the surface of the paraboloid z = 16 − x2 − y2 which is above the plane z = 0 and

D be its projection on the xy-plane

S : z = 16 − x2 − y2 

$z_x=-2x$   and $z_y=-2y$ 

Surface Area = ${\iint }_{S}dS$

= ${\iint }_D\surd (1+4x^2+4y^2)dxdy$  

=  ${\iint }_D\surd (1+4(x^2+y^2))dxdy$

Where  D : $x^2+y^2=16$   be its projection on the xy-plane 

= ${\int }_0^{2\pi }{\int }_0^4\surd 1+4r^2\cdot rdrd\theta$

Hence Option(1) and Option(4) are correct 

Explanation:

${\int }_0^{2a}{x}^3\sqrt{2ax-x^2}dx=\frac{p}{q}\pi a^5$

Let  $u=2ax-x^2$

so  $\sqrt{2ax-x^2}=\sqrt{u}$

Now, $2ax-x^2=u$  implies $du=(2a-2x)dx$  , or du = -2(x - a) dx 

Using the substitution  $u=2ax-x^2$ :

1. When x = 0 : u = 0 ,

2. When x = 2a : u = 0 (as $2ax-x^2=0$  at x = 2a ).

The limits of integration become symmetric about x = a

Rewriting x in terms of u , we evaluate the integral using symmetry : 

${\int }_0^{2a}{x}^3\sqrt{2ax-x^2}dx=\frac{103}{6}\pi a^5$   

Thus p = 103,  q = 6

$p^2+q^2={103}^2+6^2=10609+36=10645$   

Hence Option(1) is the correct answer.

Explanation:   

The Gamma function is defined as:

$\Gamma (z)={\int }_0^{\infty }{t}^{z-1}{e}^{-t}dt$

t = x² 

dt = 2x dx 

dx = dt / (2√t)

Our integral becomes:

${\int }_0^{\infty }{e}^{-x²}dx={\int }_0^{\infty }{e}^{-t}(dt/(2\surd t))$   

$=(1/2){\int }_0^{\infty }{t}^{-1/2}{e}^{-t}dt$  

This integral now matches the form of the Gamma function with z = 1/2:

$(1/2){\int }_0^{\infty }{t}^{1/2-1}{e}^{-t}dt=(1/2)\Gamma (1/2)$   

 Γ(1/2) = √π 

Therefore,  ${\int }_0^{\infty }{e}^{-x²}dx=(1/2)\surd \pi =\surd \pi /2$

Hence, option(3) is the correct answer.

Explanation:
The integral of f(x) being zero does not imply f(x) is identically zero;

f(x) may take both positive and negative values, so its areas cancel out

For example, f(x) = x - 0.5 satisfies the condition

Hence Option(3) is the correct answer.

Explanation: 

$f(x)=x^x$   as  $f(x)=e^{x\ln (x)}$ :  

$\underset{x\to 0^{+}}{lim}x\ln (x)=\underset{x\to 0^{+}}{lim}\frac{\ln (x)}{1/x}$    
  
Using L 'Hôpital's Rule:  

$\underset{x\to 0^{+}}{lim}\frac{\ln (x)}{1/x}=\underset{x\to 0^{+}}{lim}\frac{1/x}{-1/x^2}=\underset{x\to 0^{+}}{lim}-x=0$   
  
Thus,  $\underset{x\to 0^{+}}{lim}f(x)=e^0=1$  

Hence Option(2) is the correct answer.

Explanation: 

Integration of Power of Tangent functions : 

$I_n=\int ta{n}^{n}xdx$  

Reduction Formula is 

$I_n=\frac{1}{n-1}{\tan }^{n-1}(x)-I_{n-2}$   

Hence Option(1) is the correct answer.

Explanation:

Using the given reduction formula:

$I_n=-\frac{1}{n}{\sin }^{n-1}(x)\cos (x)+\frac{n-1}{n}{I}_{n-2}$ ,
  
for n = 4 ,  

$I_4=-\frac{1}{4}{\sin }^3(x)\cos (x)+\frac{3}{4}{I}_2$  

Using   ${\sin }^2(x)=\frac{1-\cos (2x)}{2}$
  
So,    $I_2=\int {\sin }^2(x)dx=\int \frac{1-\cos (2x)}{2}dx=\frac{x}{2}-\frac{\sin (2x)}{4}$   

Substitute  $I_2$  into the expression for $I_4$ :  

$I_4=-\frac{1}{4}{\sin }^3(x)\cos (x)+\frac{3}{4}(\frac{x}{2}-\frac{\sin (2x)}{4})$   

Simplify,   $I_4=-\frac{1}{4}{\sin }^3(x)\cos (x)+\frac{3x}{8}-\frac{3}{16}\sin (2x)$

Hence Option(2) is the correct answer.

Explanation:

S be that part of the surface of the paraboloid z = 16 − x2 − y2 which is above the plane z = 0 and

D be its projection on the xy-plane

S : z = 16 − x2 − y2 

$z_x=-2x$   and $z_y=-2y$ 

Surface Area = ${\iint }_{S}dS$

= ${\iint }_D\surd (1+4x^2+4y^2)dxdy$  

=  ${\iint }_D\surd (1+4(x^2+y^2))dxdy$

Where  D : $x^2+y^2=16$   be its projection on the xy-plane 

= ${\int }_0^{2\pi }{\int }_0^4\surd 1+4r^2\cdot rdrd\theta$

Hence Option(1) and Option(4) are correct 

Explanation:

(A) ${\int }_0^2{\int }_0^{x}f(x,y)dydx$ 

This indicates that f(x, y) is integrated first with respect to y , then with respect to x

The outer integral is over x from 0 to 2, and the inner integral is over y from 0 to x

⇒ x = 0 , x = 2 and y = 0 , y = x

By Changing Order of integration:

y = 0 , y = 2 and x = y , x = 2

Now, Integral becomes 

${\int }_0^2{\int }_y^{2}f(x,y)dxdy$

⇒ A → III

(B) ${\int }_0^1{\int }_y^{1}f(x,y)dxdy$  

The outer integral is over y from 0 to 1, and the inner integral is over x from y to 1

This indicates that f(x, y) is integrated first with respect to x , and the integration limit depends on y

By Changing Order of integration: 

⇒ x = 0 to 1 and y = 0 to x

Now Integrals becomes :

${\int }_0^1{\int }_0^{x}f(x,y)dydx$  

⇒ B → II

(C) ${\int }_0^2{\int }_x^{2}f(x,y)dydx$

here , x = 0 to 2 and y = x to 2

Changing order of integration:

y = 0 to 2 and x = 0 to y 

Now, Integrals becomes 

${\int }_0^2{\int }_0^{y}f(x,y)dxdy$

⇒ C → IV

(D) ${\int }_0^1{\int }_0^{y}f(x,y)dxdy$

Here y = 0 to 1 and x = 0 to y 

Changing order of integration:

here x = 0 to 1 and y = x to 1 

Now, Integrals becomes 

${\int }_0^1{\int }_x^{1}f(x,y)dydx$

⇒ D → I

Matching List-I with List-II

(A) matches (III)

(B) matches (II)

(C) matches (IV)

(D) matches (I)

Hence Option(2) is the correct answer.

Explanation:

The Graph of the region bounded by the curves y = ex and x = 1 in the first quadrant  

x = 0 and x = 1

then y = 0 and y = ex  

A =   ${\int }_{x=0}^1{\int }_{y=0}^{e^x}dxdy$   

A = e -1 

Hence Option (4) is the correct answer.

Explanation:

Let  $x=\sin t,y=\cos t$

The arc length element ds is:

$ds=\Vert {\overrightarrow{r}}^{\prime }(t)\Vert dt,$   

where,   ${\overrightarrow{r}}^{\prime }(t)=\frac{d}{dt}(\sin t,\cos t)=(\cos t,-\sin t)$

$\Vert {\overrightarrow{r}}^{\prime }(t)\Vert =\sqrt{(\cos t{)}^2+(-\sin t{)}^2}=\sqrt{{\cos }^{2}t+{\sin }^{2}t}=1$  

Thus, ds = dt 

$x^2=(\sin t{)}^2,y=\cos t$

So  $x^{2}y=(\sin t{)}^2\cos t$  

The integral becomes:

${\int }_C(1+x^{2}y)ds={\int }_0^{\frac{\pi }{2}}(1+(\sin t{)}^2\cos t)dt$  

⇒  ${\int }_0^{\frac{\pi }{2}}(1+(\sin t{)}^2\cos t)dt={\int }_0^{\frac{\pi }{2}}1dt+{\int }_0^{\frac{\pi }{2}}(\sin t{)}^2\cos tdt$

For ,  ${\int }_0^{\frac{\pi }{2}}(\sin t{)}^2\cos tdt={\int }_0^1{u}^{2}du$  

${\int }_0^1{u}^{2}du=\frac{u^3}{3}{|}_0^1=\frac{1^3}{3}-\frac{0^3}{3}=\frac{1}{3}$  

So,  ${\int }_C(1+x^{2}y)ds=\frac{\pi }{2}+\frac{1}{3}$  

Hence Option(3) is the correct answer