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

Calculation

Given \(f(0) = 0 \) & \(f'(1) = 0\) & \(f(1) = \frac{1}{2}\)

\(\therefore f(x) = \frac{2x - x^2}{2}\)

\(f^{-1}(x)\) is the image of \(f(x)\) on \(y = x.\)

Also, \(2x + 2y = 3\) passes through \(A(1, \frac{1}{2})\) & \(B(\frac{1}{2}, 1)\)

so bounded Area \(A\)

 \(= AreaOAB = 2[Area OCM + Area CMNA - Area ONA]\)

\(A = 2[\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} + \frac{1}{2} (\frac{3}{4} + \frac{1}{2}) \times \frac{1}{4} - \frac{1}{2} \int_0^1 (2x - x^2) , dx]\)

\(\Rightarrow A = 2[\frac{9}{32} + \frac{5}{32} - [x^2 - \frac{x^3}{3}]_0^1]\)

\(\Rightarrow A = 2[\frac{14}{32} -(1 - \frac{1}{3})] \)
\(​\Rightarrow A = \frac{28}{32} - \frac{2}{3} = \frac{7}{8} - \frac{2}{3} \)\(\Rightarrow A = \frac{21 - 16}{24} = \frac{5}{24} \)

⇒ 48A = 10

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:

\( \int_{\sqrt{2}}^{2} f(x) dx = \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^{2} 3 dx \)

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

\( \int_{\sqrt{2}}^{\sqrt{3}} 2 dx = 2 \times (\sqrt{3} - \sqrt{2}) \)

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

\( \int_{\sqrt{3}}^{2} 3 dx = 3 \times (2 - \sqrt{3}) \)

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

We are tasked with finding:

\( \int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor \, dx \)

Decomposing the integral based on the function:

For \\(\frac{\sqrt{3}}{2} \leq x < 1 \), x² lies between \(\frac{3}{4} \) and 1, so \(\left\lfloor x^2 \right\rfloor = 0 \). Therefore,

\( \int_{\frac{\sqrt{3}}{2}}^1 0 \, dx = 0 \)

For\( 1 \leq x < \sqrt{2} \), x²  lies between 1 and 2, so \(\left\lfloor x^2 \right\rfloor = 1 .\) Therefore,

\( \int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1 \)

For \(\sqrt{2} \leq x < \sqrt{3} \), x²  lies between 2 and 3, so \(\left\lfloor x^2 \right\rfloor = 2\) . Therefore,

\( \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2(\sqrt{3} - \sqrt{2}) \)

Summing up all the results:

\( \int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor \, dx = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) \)

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

Hence, the correct answer is Option 2. 

Calculation: 

Given,

The equation of the curve is y = 4x , and the line x = 4 intersects the curve at the point (4, 16) . We need to find the area bounded by the curve, the x-axis, and the line x = 4 .

The region of interest is a right triangle with a base along the x-axis from x = 0 to x = 4  and a height of 16 units, corresponding to the point (4, 16) .

The area of the triangle is given by the formula:

\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

Substituting the values of the base (4 units) and the height (16 units):

\( \text{Area} = \frac{1}{2} \times 4 \times 16 = 32 \, \text{square units} \)

∴ The area is 32 square units.

Hence, the correct answer is option 3.

Calculation:

Given,

The slope of the tangent to the curve y = f(x) at (x, f(x)) is 4 for every real number x , and the curve passes through the origin.

The slope of the tangent is the derivative of the function, so we have:

\( f'(x) = 4 \)

Integrating f'(x) = 4  with respect to  x :

\( f(x) = 4x + C \)

The curve passes through the origin, so when x = 0 , y = 0 . Substituting these values into the equation f(x) = 4x + C :

\( 0 = 4(0) + C \quad \Rightarrow \quad C = 0 \)

Therefore, the equation of the curve is:

\( f(x) = 4x \)

This is the equation of a straight line with a slope of 4, passing through the origin.

∴ The curve is a straight line with a slope of 4, passing through the origin.

Hence, the correct answer is option 1.

Concept:

Definite Integral properties:

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)
Calculation:

Let f(x) = x(1 – x)⁹

Now using property, \(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)

\(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}} = \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){\left\{ {1 - \left( {1 - {\rm{x}}} \right)} \right\}^9}{\rm{dx}}\)

\(⇒ \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){{\rm{x}}^9}{\rm{dx}}\)

\(⇒ \mathop \smallint \limits_0^1 \left( {{{\rm{x}}^9} - {{\rm{x}}^{10}}} \right){\rm{dx}}\)

\(⇒\left[ {\frac{{{{\rm{x}}^{10}}}}{{10}} - \frac{{{{\rm{x}}^{11}}}}{{11}}} \right]_0^1\)

⇒ 1/10 – 1/11

⇒ 1/110

∴ The value of integral \(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}}\) is 1/110.

Concept:

\(\rm \int \frac{dx}{x^2+a^2} = \frac{1}{a}\; \tan^{-1}(\frac{x}{a}) + c\)

Calculation:

Let I = \(\displaystyle\int_0^2 \rm \dfrac{dx}{x^2+4}\)

= \(\displaystyle\int_0^2 \rm \dfrac{dx}{x^2+2^2}\)

\(= \rm [\frac{1}{2}\; \tan^{-1}(\frac{x}{2})] _{0}^{2}\)

\(= \rm [\frac{1}{2}\; \tan^{-1}1 -\frac{1}{2}\; \tan^{-1}0 ]\)

\(= \rm \frac{1}{2}\times \frac{\pi}{4} - 0\)

= \(\rm \dfrac{\pi}{8}\)

Concept:

The area under the curve y = f(x) between x = a and x = b, is given by:

Area = \(\rm\int_{a}^{b}ydx\)

Similarly, the area under the curve y = f(x) between y = a and y = b, is given by:

Area = \(\rm\int_{a}^{b}xdy\)

Calculation:

Here, 

x² = y  and line y = 1 cut the parabola

∴ x² = 1

⇒ x = 1 and -1

\( \text{Area =}\int_{-1}^{1} y d x \)

Here, the area is symmetric about the y-axis, we can find the area on one side and then multiply it by 2, we will get the area,

\( \text{Area}_1 = \int_{0}^{1} y d x \)

\( \text{Area}_1 = \int_{0}^{1} x^{2} dx \)

\(= \rm\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{3}\)

This area is between y = x² and the positive x-axis.

To get the area of the shaded region, we have to subtract this area from the area of square i.e.

\((1 \times 1)-\frac{1}{3}=\frac{2}{3}\)

\(Total\;Area=2\times{\frac{2}{3}} =\frac{4}{3}\) square units.

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation: 

I = \(\rm \int_0^1 x \sqrt{x^2 + 4}\;dx\)

Let x² + 4 = t

Differentiating with respect to x, we get

⇒ 2xdx = dt

⇒ xdx = \(\rm \frac {dt}{2}\)

Now,

I = \(\rm \frac{1}{2}\int_4^5 \sqrt{t}\;dt\)

= \(\rm \frac{1}{2} \left[\frac{t^{3/2}}{\frac{3}{2}} \right ]_4^5\)

= \(\rm \frac{1}{3} \left[5^{3/2} - 4^{3/2} \right ]\)

= \(\rm \frac{1}{3}[5\sqrt 5 - 8]\)

Concept:

1 + cos 2x = 2cos² x

1 - cos 2x = 2sin² x

\(\rm \int \cos x\;dx = \sin x + c\)

Calculation:

I = \(\rm \int cos^2 x\;dx\)

= \(\rm \int \frac{1+\cos 2x}{2}\;dx\)

= \(\rm \frac{1}{2}\int (1+\cos 2x)\;dx\)

= \(\rm \frac{1}{2} \left[x+\frac{\sin 2x}{2} \right ] + c\)

= \(\rm \frac{x}{2}+\frac{\sin 2x}{4} + c\)

Concept:

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{dx}}\)

Calculation: 

Let I = \(\rm \int _{0}^{2π} \frac{\sin 2x}{a -b\cos x}dx\)         ----(1)

Using property f(a + b – x),

I = \(\rm \int _{0}^{2π} \frac{\sin 2(2π -x)}{a -b\cos (2π -x) }dx\)   

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

I = \(\rm \int _{0}^{2π} \frac{-\sin 2x}{a -b\cos x}dx\)         ----(2)       

I = -I

2I = 0

∴ I = 0

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation: 

I = \(\rm \int_{1}^{\infty} \frac{4}{x^4}dx\)

= \(\rm \int_{1}^{\infty}4{x^{-4}}dx\)

= \(\rm \left[\frac{4x^{-3}}{-3} \right ]_1^{\infty}\)

= \(\rm \frac{-4}{3}\left[\frac{1}{x^3} \right ]_1^{\infty}\)

= \(\rm \frac{-4}{3}\left[\frac{1}{\infty} - \frac{1}{1}\right ]\)

= \(\rm \frac{-4}{3}[0-1]\)

= \(\frac 4 3\)

Concept:

\(\rm \displaystyle\int_{a}^{b} f(x) dx = \displaystyle\int_{a}^{b} f(a+b-x) dx\)

Calculations:

Consider, I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\)             ....(1)

I = \(\rm \displaystyle\int_{0}^{\pi/2} \dfrac{\sqrt{\sin (\dfrac{\pi}{2}-x)}}{\sqrt{\sin (\dfrac{\pi}{2}-x)}+ \sqrt{\cos (\dfrac{\pi}{2}-x)}}dx\)

I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx\)                           ....(2)

Adding (1) and (2), we have

2I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos x}+ \sqrt{sinx}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx\)

2I = \(\rm \displaystyle\int_0^{\pi/2}dx\)

2I = \(\rm[x]^\frac{\pi}{2}_0\)

I = \(\dfrac{\pi}{4}\)

Concept: 

\(\int√{a^2-x^2}\;dx=\frac{x}{2} {√{x^2-a^2}}+\frac{a^2}{2}sin^{-1}\frac{x}{a}+c\) 

Function y = √f(x) is defined for f(x) ≥ 0. Therefore y can not be negative.

Calculation:

Given: 

y = \(\rm √{16-x^2}\) and x-axis

At x-axis, y will be zero

y = \(\rm √{16-x^2}\)

⇒ 0 = \(\rm √{16-x^2}\)

⇒ 16 - x² = 0

⇒ x2 = 16

∴ x = ± 4

So, the intersection points are (4, 0) and (−4, 0)

Since the curve is y = \(\rm √{16-x^2}\)

So, y ≥ o [always]

So, we will take the circular part which is above the x-axis

Area of the curve, A \(\rm =\int_{-4}^{4}√{16-x^2}\;dx\)

We know that,

\(\int√{a^2-x^2}\;dx=\frac{x}{2} {√{x^2-a^2}}+\frac{a^2}{2}sin^{-1}\frac{x}{a}+c\)

= \( \rm [ \frac x 2 √{(4^2- x^2) }+ \frac {16}{2}sin^{-1} \frac x4]_{-4}^{4 }\) 

= \( \rm [ \frac x 2 √{(4^2- 4^2) }+ \frac {16}{2}sin^{-1} \frac 44]- \rm [ \frac x 2 √{(4^2- (-4)^2) }+ \frac {16}{2}sin^{-1} \frac {4}{-4})]\)

= 8 sin^-1 (1) + 8 sin-1 (1)

= 16 sin^-1 (1)

= 16 × π/2

= 8π sq units

Concept:

\(\rm \int \frac{1}{\sqrt{a^2-x^2}}dx= \sin^{-1 } \left(\frac{x}{a} \right ) + c\)

Calculation:

I = \(\rm \int \frac {1}{\sqrt{16-25x^2}}dx\)

= \(\rm \int \frac {1}{\sqrt{16-(5x)^2}}dx\)

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = \(\rm \frac {dt}{5}\)

Now,

I = \(\rm \frac {1}{5}\int \frac {1}{\sqrt{4^2-t^2}} dt\)

= \(\rm \frac 1 5 \sin^{-1} \left(\frac t 4 \right)\) + c

= \(\rm \frac 1 5 \sin^{-1} \left(\frac {5x} {4} \right)\) + c