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

Concept:

The area under the curve y = f(x) between x = a and x = b,is given by,  Area = \(\rm \int_{x=a}^{x =b}f(x) \;dx\)

\(\smallint {{\rm{x}}^{\rm{n}}}{\rm{dx}} = \frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}} + {\rm{C}}\)

Calculation:

Here, we have to find the area of the region bounded by the curves y = \(\rm \frac{x^2}{2}\), the line x = 2, x  = 0 and the x - axis

So, the area enclosed by the given curves = \(\rm \int_0^2 {\rm \frac{x^2}{2}}\;dx\)

As we know that, \(\smallint {{\rm{x}}^{\rm{n}}}{\rm{dx}} = \frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}} + {\rm{C}}\)

\(=\rm \int_0^2 {\frac {x^2}{2}}\;dx = \left[ {\frac{{{x^3}}}{6}} \right]_0^2\)

\(\rm = \frac{1}{6}\;\left( {8- 0\;} \right) = \frac 4 3 \;sq.\;units\)

Hence, option 4 is the correct answer.

Calculation

Area = \(\int_{0}^{\pi} |\cos x| dx\)

cos x is positive from 0 to π/2 and negative from π/2 to π.

So, we split the integral:

Area = \(\int_{0}^{\frac{\pi}{2}} \cos x dx + \int_{\frac{\pi}{2}}^{\pi} (-\cos x) dx\)

Area = \(\left[ \sin x \right]_{0}^{\frac{\pi}{2}} - \left[ \sin x \right]_{\frac{\pi}{2}}^{\pi}\)

Area = \(\left( \sin \frac{\pi}{2} - \sin 0 \right) - \left( \sin \pi - \sin \frac{\pi}{2} \right)\)

Area = \((1 - 0) - (0 - 1)\)

Area = \(1 - (-1)\)

Area = \(1 + 1\)

Area = \(2\)

∴ The area bounded by the curve y = cos x, x = 0 and x = π is 2.

Hence option 1 is correct

Calculation:

Find the equation of line first intercept of y is 2

equation becomes:

y = x + 2

If intercept of y is n

Equation becomes

y = x + n

I = \(\mathop \smallint \limits_1^3 {y^2}dx\)

I = \(\mathop \smallint \limits_1^3 {\left( {x + n} \right)^2}dx\)

I = \(\mathop \smallint \limits_1^3 \left[ {{{\rm{x}}^2} + {{\rm{n}}^2} + 2{\rm{xn}}} \right]{\rm{dx}}\)

n = 2

I = \(\mathop \smallint \limits_1^3 \left( {{{\rm{x}}^2} + 4 + 4{\rm{x}}} \right){\rm{dx}}\)

I = \(\left[ {\frac{{{{\rm{x}}^2}}}{3} + 4x + \frac{{4{{\rm{x}}^2}}}{3}} \right]_1^3\)

\(\begin{array}{l} {\rm{I}} = \left[ {\frac{{{3^3}}}{3} + 12 + 18 - \frac{1}{3} - 4 - 2} \right]\\ {\rm{I}} = 33 - \frac{1}{3} \end{array}\)

Calculation

\(\cos x > \sin x, \forall x \in \left( 0, \frac{\pi}{4} \right)\) and \(\cos x < \sin x, \forall x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right)\)

\(y_{1} = \sin x + \cos x\)

\(y_{2} = \left| \cos x - \sin x \right|\)

\(\Rightarrow \text{Area}\)

\(\int_{0}^{\pi/2} \left( y_{1} - y_{2} \right) dx\)

\(\int_{0}^{\pi/4} \left( (\sin x + \cos x) - (\cos x - \sin x) \right) dx + \int_{\pi/4}^{\pi/2} \left( (\sin x + \cos x) - (\sin x - \cos x) \right) dx\)

\(= 4 - 2\sqrt{2}\)

Hence option 2 is correct

Calculation

Given:

Parabola: y² = 4x

Line: x = 3

Since y² = 4x, then y = \(2\sqrt{x}\) (in the first quadrant, y > 0)

Required Area = 2 × (Area of the region OCAO)

⇒ Area = \(2 \int_{0}^{3} y \, dx\)

⇒ Area = \(2 \int_{0}^{3} 2\sqrt{x} \, dx\)

⇒ Area = \(4 \int_{0}^{3} x^{\frac{1}{2}} \, dx\)

⇒ Area = \(4 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{3}\)

⇒ Area = \(4 \times \frac{2}{3} \left[ (3)^{\frac{3}{2}} - 0 \right]\)

⇒ Area = \(\frac{8}{3} (3\sqrt{3})\)

∴ Required area = \(8\sqrt{3}\) sq. units.

Hence option 2 is correct

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: 

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

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

Calculation:

Enclosed Area

\( = 2\mathop \smallint \nolimits_0^{\pi /4} \left( {\cos x - \sin x} \right)dx\)

\( = 2\left[ {\sin x + \cos x} \right]_0^{\pi /4}\)

\( = 2\left[ {\left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right) - \left( {0 + 1} \right)} \right]\)

\( = 2\left( {\sqrt 2 - 1} \right)\)

Concept:

Area under a Curve by Integration

Find the area under this curve by summing vertically.

• In this case, we find the area is the sum of the rectangles, height x = f(y) and width dy.
• If we are given y = f(x), then we need to re-express this as x = f(y) and we need to sum from the bottom to top.

So, \({\bf{A}} = \mathop \smallint \nolimits_{\bf{a}}^{\bf{b}} {\bf{xdy}} = \mathop \smallint \nolimits_{\bf{a}}^{\bf{b}} {\bf{f}}\left( {\bf{y}} \right){\bf{dy}}\)

Calculation:

Given Curve: x = 4 - y2

⇒ y2 = 4 - x
⇒ y2 = - (x - 4)           

The above curve is the equation of the Parabola,

We know that at y-axis; x = 0

⇒ y2 = 4 - x

⇒ y2 = 4 - 0 = 4

⇒ y = ± 2

⇒ (0, 2) or (0, -2) are Point of intersection.

Area under the curve \( = \mathop \int \nolimits_{-2}^2 {\rm{xdy}}\)

\(= \rm \int_{-2}^2 (4-y^2)dy\)

\(\rm = \left[4y- {\frac{{{{ {{\rm{y}} } }^3}}}{3}} \right]_{-2}^2\)

\(= \frac{32}{3}\) Sq. unit

Explanation:

Equation of circle is given by x² + y² = a²

Let's take the strip along a y-direction and integrate it from 0 to 'a' this will give the area of the first quadrant and in order to find out the area of a circle multiply by 4

\(y = \sqrt {x^2 - a^2}\)

Area of first Quadrant = \(\mathop \smallint \limits_0^a y\;dx\) = \(\mathop \smallint \limits_0^a \sqrt {{a^2} - {x^2}} \;dx\)

Area of circle = 4 × \(\mathop \smallint \limits_0^a \sqrt {{a^2} - {x^2}} \;dx\)

Given:

The parabolas y = 3x2 and x2 - y + 4 = 0

Concept:

Apply concept of area between two curves y₁ and y₂ between x = a and x = b

\(\rm A=\int_a^b(y_1-y_2)\ dx\)

Calculation:

The parabolas y = 3x2 and x2 - y + 4 = 0

then 3x² = x² + 4

⇒ x² = 2

⇒ x = ± √ 2

Then the area is 

\(\rm A=\int_{-\sqrt2}^{\sqrt2}(x^2+4-3x^2) \ dx\)

\(\rm A=\int_{-\sqrt2}^{\sqrt2}(4-2x^2) \ dx\)

\(\rm A=[4x-2\frac{x^3}{3}]_{-\sqrt2}^{\sqrt2}\)

\(\rm A=4[\sqrt2-(-\sqrt2)]-\frac{2}{3}[{\sqrt2}^3-{(-\sqrt2)}^3]\)

\(\rm A=8\sqrt2-\frac{8}{3}\sqrt2\)

\(\rm A=\frac{16\sqrt2}{3}\) sq unit.

Hence option (4) is correct.

Concept:

The area of the curve y = f(x) is given by:

A = \(\rm \int_{x_1}^{x_2}f(x) dx\)

where x1 and x2 are the endpoints between which the area is required.

Imp. Note: The net area will be the addition of the area below the x-axis and the area above the x-axis.

Calculation:

The f(x) = y = 4x³

Given the end points x1 = -2, x2 = 3

Area of the curve (A) = \(\rm \left|\int_{-2}^3 4x^3dx\right|\)

⇒ A = \(\rm \left|\int_{-2}^0 4x^3dx\right| + \left|\int_0^3 4x^3dx\right|\)

⇒ A = \(\rm \left|4\left[x^4\over4\right]_{-2}^0\right| + \left|4\left[x^4\over4\right]_0^3\right|\)

⇒ A = \(\rm \left|\left[0- 2^4\right]\right| + \left|\left[3^4 - 0\right]\right| \)

⇒ A = \(\rm \left|-16\right| + \left|81\right|\)

⇒ A = 97

Additional Information

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C 

Explanation:

Given equation of curves are

y = x²    ---(1) and y = 16    ---(2)

By solving both equation (1) and (2) we have:

x² = 16

x = 4, -4.

∴ Points of intersection are (4, 16) and (-4, 16).

From the figure we have,

\(Required~Area~=~∫_{-4}^4(16-x^2)~dx \)

By using Integral property we have,

 \(A=~2∫_{0}^4(16-x^2)~dx \)

\(= 2\left[ {16x - \frac{{{x^3}}}{3}} \right]_0^4\)

\(= 2\left[ {16x - \frac{{{x^3}}}{3}} \right]_0^4\)

\( = 2\left[ {64 - \frac{{64}}{3}} \right] \)

\(= 2 \times 64 \times \frac{2}{3}\;\)

 \(A=\frac{256}{3}~sq.units\)

Alternate Method 

There is another method also by which we can solved the problem,

By considering horizontal strip and by the condition of symmetry we have:

\(Area~=~2\int_0^{16}x~dy\)

\(Area~=~2\int_0^{16}\sqrt{y}~dy\)

\(Area~=~2~\times~\frac{2}{3}~\times~[y^{\frac{3}{2}}]_0^{16} \)

\(Area~=~2\times \frac{2}{3}\times [16^{\frac{3}{2}}-0]\)

Area = \(\frac{256}{3}~sq.unit\)

Concept:

The area under a Curve by Integration:

Find the area under this curve is by summing horizontally.

In this case, we find the area is the sum of the rectangles, heights y = f(x) and width dx.

We need to sum from left to right.

∴ Area =  \( \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{ydx}} = {\rm{\;}}\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

Calculation: 

Here, we have to find the area of the region bounded by the curves y = x2, x-axis and ordinates x = -1 and x = 2

So, the area enclosed by the given curves is given by \(\rm \mathop \int \nolimits_{-1}^2{x^2}\;dx\)

As we know that, \(\smallint {{\rm{x}}^{\rm{n}}}{\rm{dx}} = \frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}} + {\rm{C}}\)

Area = \(\rm \mathop \int \nolimits_{-1}^2{x^2}\;dx\)

= \( \rm \left[ {\frac{{{x^3}}}{3}} \right]_{-1}^2\)

= \(\left[\frac 83 - \frac {-1}{3}\right] = \frac 93=3\)

Area = 3 sq. units.

Explanation:

Given curves are y = x - 1 and y2 = 2x + 6 

On solving, we get, 

y² = 2(y + 1) + 6

⇒ y² - 2y - 8 = 0

⇒ (y - 4)(y + 2) = 0

⇒ y = -2, 4

Now, we can find the area by

A = \(\int_{-2}^{4}\left [y+1-\left ( \frac{y^{2}}{2}-3 \right ) \right ]dy \)

= \(\int_{-2}^{4}\left (4+y-\frac{y^{2}}{2} \right )dy \)

= \(\left [ 4y+\frac{y^{2}}{2} -\frac{y^{3}}{6}\right ]_{-2}^{4}\)

= \(16+8-\frac{32}{3}-\left ( -8+2+\frac{4}{3} \right )\)

∴ A = 18