Application of Integrals MCQ Quiz in मल्याळम - Objective Question with Answer for Application of Integrals - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculations:

To find the area of the curve y = sin x, lying between x = 0, y = 0 and x = 2π, first draw the graph of the given curve.

Required area = area of OABO + area of BCDB.

= area of OABO  + area of OABO 

= 2 (area of OABO) 

= 2  \(\rm \int_{0}^{\pi}sinxdx\)

= 2\(\rm [-\;cos\;x]_{0}^{\pi}\)

= - 2 [\(\rm cos\;\pi - cos \;0\)]

=  - 2(-1 - 1)

= 4 

Concept:

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

Calculation:

Equation of curve ⇒ y = x²

Equation of line ⇒ y = 2x + 3

Line intersect the curve

\(\therefore {\rm{\;}}{{\rm{x}}^2}{\rm{\;}} = {\rm{\;}}2{\rm{x\;}} + {\rm{\;}}3\)

\(\Rightarrow {{\rm{x}}^2} - 2{\rm{x}} - 3 = 0\)

\(\Rightarrow {{\rm{x}}^2} - 3{\rm{x}} + {\rm{x}} - 3 = 0\)

\(\Rightarrow {\rm{x}}\left( {{\rm{x}} - 3} \right) + 1\left( {{\rm{x}} - 3} \right) = 0\)

\(\Rightarrow \left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 1} \right) = 0 \Rightarrow {\rm{x}} = 3,{\rm{\;}} - 1\)

So the line cut the curve at x = 3 and x = -1

Now area of the region bounded by curve = area below the line – area below curve

\(\therefore {\rm{Area}} = \mathop \smallint \nolimits_{ - 1}^3 \left( {2{\rm{x}} + 3 - {{\rm{x}}^2}} \right)\)

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

\(= \left[ {9 + 9 - 9 - 1 + 3 - \frac{1}{3}} \right]\)  

\(= \left[ {8 + \frac{8}{3}} \right]\)

\(= \frac{{32}}{3}{\rm{\;sq\;units}}\)

Hence, option (2) is correct.

Concept:

Area bounded by function f(x) and g(x) is given as,

Area \(= \left| {\mathop \smallint \limits_{{{\rm{x}}_1}}^{{{\rm{x}}_2}} \left\{ {{\rm{f}}\left( {\rm{x}} \right) - {\rm{g}}\left( {\rm{x}} \right)} \right\}{\rm{dx}}} \right|\).

To find x₁ and x₂ we need to solve for f(x) = g(x).

Calculation:

Given: f(x) = e^-2x and g(x) = 0

x₁ = -1 and x₂ = 1

Plot a rough graph of given curves and shade the bounded area to be calculated.

So, area under the curve \(= \left| {\mathop \smallint \limits_{{x_1}}^{{x_2}} \left\{ {f\left( x \right) - g\left( x \right)} \right\}dx} \right|\)

\(= \left| {\mathop \smallint \limits_{ - 1}^1 \left\{ {{e^{ - 2x}} - 0} \right\}dx} \right|\)

\(= \left| { - \frac{{{e^{ - 2x}}}}{2}} \right|_{ - 1}^1\)

\(= \frac{{ - {{\rm{e}}^{ - 2{}}}{\rm{\;}} - {\rm{\;}}\left( { - {{\rm{e}}^{2}}} \right)}}{2}\)

\(= \frac{{{{\rm{e}}^{2{}}}{\rm{\;}} - {\rm{\;}}{{\rm{e}}^{ - 2{}}}}}{2}\)

Concept:

Area under a curve:

• The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral \(\rm \left|\int_a^b f(x)\ dx\right|\), for curves which are entirely on the same side of the x-axis in the given range.
• If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.

Calculation:

The graph of the curve y = 2a sin x is presented 

∴ The required area = \(\rm \left|\int_0^{\pi} 2a\sin x\ dx\right|\) = 2a \(\rm \left[-\cos x\right]_0^{\pi}\) =  4a.

Calculation:

The area bounded by the curves is y2 = 4x,

\(\int_{4}^{9}\sqrt{4x}dx\)

\(\Rightarrow 2\int_{4}^{9}\sqrt{x}dx\)

\(\Rightarrow 2\left ( \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right )_{4}^{9}\)

\(\Rightarrow 2\left ( \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right )_{4}^{9}\)

\(\Rightarrow 2\times \frac{2}{3}\left ( 9^{\frac{3}{2}}-4^{\frac{3}{2}} \right )\)

\(\Rightarrow \frac{4}{3}\left ( 3^{3}-2^{3} \right )\)

\(\Rightarrow \frac{4}{3}\left ( 27-8 \right )\)

\(\Rightarrow \frac{4}{3}\times 19\)

\(\Rightarrow \frac{76}{3}\)

Calculation:

Given:  \(\rm y=x\) and \(\rm y=2x+6\).

Finding a point of intersection:

\(\Rightarrow \rm x= 2x+6\Rightarrow x=-6 \)

Thus, y = - 6.

Draw the graph of the curve  \(\rm y=x\) and \(\rm y=2x+6\).

Let the shaded area be A.

Using the formula of the area under the curve, \(\rm A=\left | \int_{a}^{b}f(x)-g(x) dx \right |\)

\(\Rightarrow \rm A=\left | \int_{-6}^{0}(2x+6-x) dx \right |=\left | \int_{-6}^{0}(x+6) dx \right |=\left | \left [ \frac {x^{2}}{2}+6x\right ]_{-6}^{0} \right |\)

Substitute the limit to evaluate the area.

\(\Rightarrow \rm A=\left | {0+0-\frac{36}{2}} +36 \right |=18\)

Hence, the required answer is option 2.

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 = x3, the line x = 2, x  = 5 and the x-axis

So, the area enclosed by the given curves is given by \(\rm \mathop \smallint \nolimits_1^2 {x^3}\;dx\)

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

Area = \(\rm \mathop \smallint \nolimits_1^2 {x^3}\;dx = \left[ {\frac{{{x^4}}}{4}} \right]_1^2\)

\( = \frac{1}{4}\;\left( {16 - 1\;} \right) = \frac {15}{4}\) 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 =  \( \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{ydx}} = {\rm{\;}}\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

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

Calculation: 

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

So, the area enclosed by the given curves is given by:

\(A=\rm \mathop \int \nolimits_1^2{x^2}\;dx\)

Area = \(\rm \mathop \int \nolimits_1^2{x^2}\;dx\)

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

= \(\frac 83 - \frac 13 = \frac 73\)sq. unit

Concept:

Area between Two Curves: Let curves are f(x) and g(x)

\({\rm{Area}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} \left[ {{\rm{f}}\left( {\rm{x}} \right) - {\rm{g}}\left( {\rm{x}} \right)} \right]{\rm{dx}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} \left[ {{\rm{Top}} - {\rm{bottom}}} \right]{\rm{dx}}\)

Calculation:

Given:

Equation of the parabola is y = x² and the equation of line is y = 2x + 3

Substitute the value of y in parabola equation, we get

⇒ 2x + 3 = x²

⇒ x² – 2x - 3 = 0

⇒ x² – 3x + x - 3= 0

⇒ x(x - 3) + 1(x – 3) = 0

⇒ (x + 1) (x - 3) = 0

∴ x = -1, 3

Put the value of x in y = x²

The point of intersection of the given curves are (-1, 1) and (3, 9)

Now,

Area bounded by curve \(= \mathop \smallint \nolimits_{ - 1}^3 \left[ {{\rm{Top}} - {\rm{bottom}}} \right]{\rm{dx}} = \mathop \smallint \nolimits_{ - 1}^3 \left[ {\left( {2{\rm{x}} + 3} \right) - {{\rm{x}}^2}} \right]{\rm{dx\;}}\)

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

\(= \left[ {\left( {9 + 9} \right) - \left( {1 - 3} \right)} \right] - \left( {9 - \frac{-1}{3}} \right) = 20 - \frac{{28}}{3} = \frac{{32}}{3}{\rm{sq}}.{\rm{\;unit}}\)

Concept:

Area of triangle = \(\rm \frac{1}{2}\times base \times height\)

Calculations:

Given lines are y = x, y = 0 and x = 4

To find the area of region bounded by line y = x, y = 0 and x = 4, fist draw a graph of the lines.

First find the point of intersection.

when y = 0 , x = 0

when x = 4, y = 4

So, point of intersection is (0, 0), and (4, 4).

By these lines y = x, y = 0 and x = 4, we get bounded region as triangle.

Area of triangle = \(\rm \frac{1}{2}\times base \times height\)

⇒Area of triangle = \(\rm \frac{1}{2}\times 4 \times 4\)

⇒Area of triangle = 8 units

Hence,the area of the region bounded by the lines y = x, y = 0 and x = 4 is 8 units