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  ${\int }_0^{\pi }\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}\mathrm{d}\mathrm{x}$

= 2$[-\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}{]}_0^{\pi }$

= - 2 [$\mathrm{c}\mathrm{o}\mathrm{s}\pi -\mathrm{c}\mathrm{o}\mathrm{s}0$]

=  - 2(-1 - 1)

= 4 

Concept:

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

Calculation:

Equation of curve ⇒ y = x²

Equation of line ⇒ y = 2x + 3

Line intersect the curve

$\therefore {\mathrm{x}}^2=2\mathrm{x}+3$

$\Rightarrow {\mathrm{x}}^2-2\mathrm{x}-3=0$

$\Rightarrow {\mathrm{x}}^2-3\mathrm{x}+\mathrm{x}-3=0$

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

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

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

$\therefore \mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}={\int }_{-1}^3\left(2\mathrm{x}+3-{\mathrm{x}}^2\right)$

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

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

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

$=\frac{32}{3}\mathrm{s}\mathrm{q}\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$

Hence, option (2) is correct.

Concept:

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

Area $=\left|\underset{{\mathrm{x}}_1}{\overset{{\mathrm{x}}_2}{\int }}\left\{\mathrm{f}\left(\mathrm{x}\right)-\mathrm{g}\left(\mathrm{x}\right)\right\}\mathrm{d}\mathrm{x}\right|$.

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 $=\left|\underset{x_1}{\overset{x_2}{\int }}\left\{f\left(x\right)-g\left(x\right)\right\}dx\right|$

$=\left|\underset{-1}{\overset{1}{\int }}\left\{e^{-2x}-0\right\}dx\right|$

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

$=\frac{-{\mathrm{e}}^{-2}-\left(-{\mathrm{e}}^2\right)}{2}$

$=\frac{{\mathrm{e}}^2-{\mathrm{e}}^{-2}}{2}$

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 $|{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}|$, 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 = $|{\int }_0^{\pi }2\mathrm{a}\sin \mathrm{x}\mathrm{d}\mathrm{x}|$ = 2a ${[-\cos \mathrm{x}]}_0^{\pi }$ =  4a.

Calculation:

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

${\int }_4^9\sqrt{4x}dx$

$\Rightarrow 2{\int }_4^9\sqrt{x}dx$

$\Rightarrow 2{\left(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)}_4^9$

$\Rightarrow 2{\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)}_4^9$

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

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

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

$\Rightarrow \frac{4}{3}\times 19$

$\Rightarrow \frac{76}{3}$

Calculation:

Given:  $\mathrm{y}=\mathrm{x}$ and $\mathrm{y}=2\mathrm{x}+6$.

Finding a point of intersection:

$\Rightarrow \mathrm{x}=2\mathrm{x}+6\Rightarrow \mathrm{x}=-6$

Thus, y = - 6.

Draw the graph of the curve  $\mathrm{y}=\mathrm{x}$ and $\mathrm{y}=2\mathrm{x}+6$.

Let the shaded area be A.

Using the formula of the area under the curve, $\mathrm{A}=|{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})\mathrm{d}\mathrm{x}|$

$\Rightarrow \mathrm{A}=|{\int }_{-6}^0(2\mathrm{x}+6-\mathrm{x})\mathrm{d}\mathrm{x}|=|{\int }_{-6}^0(\mathrm{x}+6)\mathrm{d}\mathrm{x}|=\left|{[\frac{{\mathrm{x}}^2}{2}+6\mathrm{x}]}_{-6}^0\right|$

Substitute the limit to evaluate the area.

$\Rightarrow \mathrm{A}=|0+0-\frac{36}{2}+36|=18$

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

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 ${\int }_1^2{\mathrm{x}}^3\mathrm{d}\mathrm{x}$

As we know that, $\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}$

Area = ${\int }_1^2{\mathrm{x}}^3\mathrm{d}\mathrm{x}={\left[\frac{{\mathrm{x}}^4}{4}\right]}_1^2$

$=\frac{1}{4}\left(16-1\right)=\frac{15}{4}$ 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 =  ${\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{y}\mathrm{d}\mathrm{x}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}$

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

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:

$A={\int }_1^2{\mathrm{x}}^2\mathrm{d}\mathrm{x}$

Area = ${\int }_1^2{\mathrm{x}}^2\mathrm{d}\mathrm{x}$

= ${\left[\frac{{\mathrm{x}}^3}{3}\right]}_1^2$

= $\frac{8}{3}-\frac{1}{3}=\frac{7}{3}$sq. unit

Concept:

Area of triangle = $\frac{1}{2}\times \mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}\times \mathrm{h}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}$

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 = $\frac{1}{2}\times \mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}\times \mathrm{h}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}$

⇒Area of triangle = $\frac{1}{2}\times 4\times 4$

⇒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

Concept:

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

$\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}={\int }_{\mathrm{a}}^{\mathrm{b}}\left[\mathrm{f}\left(\mathrm{x}\right)-\mathrm{g}\left(\mathrm{x}\right)\right]\mathrm{d}\mathrm{x}={\int }_{\mathrm{a}}^{\mathrm{b}}\left[\mathrm{T}\mathrm{o}\mathrm{p}-\mathrm{b}\mathrm{o}\mathrm{t}\mathrm{t}\mathrm{o}\mathrm{m}\right]\mathrm{d}\mathrm{x}$

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 $={\int }_{-1}^3\left[\mathrm{T}\mathrm{o}\mathrm{p}-\mathrm{b}\mathrm{o}\mathrm{t}\mathrm{t}\mathrm{o}\mathrm{m}\right]\mathrm{d}\mathrm{x}={\int }_{-1}^3\left[\left(2\mathrm{x}+3\right)-{\mathrm{x}}^2\right]\mathrm{d}\mathrm{x}$

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

$=\left[\left(9+9\right)-\left(1-3\right)\right]-\left(9-\frac{-1}{3}\right)=20-\frac{28}{3}=\frac{32}{3}\mathrm{s}\mathrm{q}.\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}$