Concept:

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

A = ${\displaystyle {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}}$

Calculation:

Here, the curve x = f(y) and lines y = a and y = b

∴Area = ${\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{y})\mathrm{d}\mathrm{y}\cdots (\text{function is f(y)})$

= ${\displaystyle {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}\mathrm{d}\mathrm{y}}$                   

(∵ f(y) = x)

Hence, option (3) is correct. 

Concept:

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

A = ${\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}$

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

Calculation:

The f(x) = y = 2e4x 

Given the end points x₁ = 0, x₂ = 4

Area of the curve (A) = |${\int }_0^4$2e4x dx|

⇒ A = $\left|{\left[\frac{2{\mathrm{e}}^{4\mathrm{x}}}{4}\right]}_0^4\right|$

⇒ A = $\frac{1}{2}\left|{\left[{\mathrm{e}}^{4\mathrm{x}}\right]}_0^4\right|$

⇒ A = $\frac{1}{2}\left|[{\mathrm{e}}^{16}-{\mathrm{e}}^0]\right|$

⇒ A = $\frac{\mathbf{1}}{\mathbf{2}}({\mathbf{e}}^{\mathbf{16}}\mathbf{-}\mathbf{1})$

Additional Information

Integral property:

• ∫ xn dx = $\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}$+ C ; n ≠ -1
• $\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\ln \mathrm{x}$ + 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 

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

Formula used:

Area of the curve y = f(x) bounded b/w x = a to x = b is  given by

$A={\int }_a^{b}f(x)dx$

Calculation:

We have 

f(x) = 2x - x²    -----(1)

y = -x              -----(2)

From equation (1)

⇒ -x = 2x - x²

⇒ x² - 3x = 0

⇒ x(x - 3) = 0

⇒ x = 0 and 3

Put these values in equation (2)

⇒ y = 0 and -3

So, parabola f(x) = 2x - x2 & line y = -x intersect at (0, 0) & (3, -3).

Hence, the Required area (shaded region)

A = ${\int }_0^3[(2\mathrm{x}-{\mathrm{x}}^2)-(-\mathrm{x})]\mathrm{d}\mathrm{x}$

⇒ A $={\int }_0^3(3\mathrm{x}-{\mathrm{x}}^2)\mathrm{d}\mathrm{x}={[\frac{3{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^3}{3}]}_0^3$

⇒ A $=\frac{27}{2}-\frac{27}{3}=\frac{9}{2}$ sq units.

Concept:

The area between the curves y1 = f(x) and y2 = g(x) is given by:

Area enclosed = ${\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}({\mathrm{y}}_1-{\mathrm{y}}_2)\mathrm{d}\mathrm{x}$

Where, x1 and x2 are the intersections of curves y1 and y2 

Calculation:

Shaded area has to be calculated

Curve 1: y = |x - 2|

⇒ y = 2 - x for x < 2

⇒ y = x - 2 for x ≥ 2

Curve 2: y = 5 - |x+1|

⇒ y = 5 + x + 1 = x + 6 for x < -1

⇒ y = 5 - x - 1 = 4 - x for x ≥ -1

Area enclosed (A) = ${\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}({\mathrm{y}}_1-{\mathrm{y}}_2)\mathrm{d}\mathrm{x}$

⇒ A = ${\int }_{-1}^2$ (5 - |x + 1|) - (|x - 2|) dx

⇒ A = $|{\int }_{-2}^{-1}\mathrm{x}+6-(2-\mathrm{x})\mathrm{d}\mathrm{x}|+|{\int }_{-1}^{2}4-\mathrm{x}-(2-\mathrm{x})\mathrm{d}\mathrm{x}|+|{\int }_2^{3}4-\mathrm{x}-(\mathrm{x}-2)\mathrm{d}\mathrm{x}|$ 

⇒ A = $|{\int }_{-2}^{-1}(2\mathrm{x}+4)\mathrm{d}\mathrm{x}|+\left|{\int }_{-1}^{2}2\mathrm{d}\mathrm{x}\right|+|{\int }_2^3(6-2\mathrm{x})\mathrm{d}\mathrm{x}|$ 

⇒ A = $|[{\mathrm{x}}^2+4\mathrm{x}{]}_{-2}^{-1}|+|[2\mathrm{x}{]}_{-1}^2|+|[6\mathrm{x}-{\mathrm{x}}^2{]}_2^3|$ 

⇒ A = $|[-3+4(1)]|+|[2(3)]|+|[6(1)-(5)]|$

⇒ A = 1 + 6 + 1 = 8 sq. units

Given:

Area bounded by curve y2 = x and the line x = 1, x = 4 and x-axis

Formula used:

$\int x^n.dx={\displaystyle \frac{x^{n+1}}{n+1}};$ n can't be - 1

Area under a curve y = f(x) is given by $\int x.dx$

Calculation:

From the figure, area is in the curve y² = x, from x = 1 to x = 4

y = x^1/2

The area is given by:

${\int }_1^4{x}^{1/2}.dx$

⇒ $|{\displaystyle \frac{x^{3/2}}{3/2}}|$, x from 1 to 4; By putting values of x:

⇒ |8/(3/2) - 1/(3/2)|

⇒ 14/3

∴ Area of the region = 14/3

Formula used : 

$\int \surd a^2-x^{2}dx=\frac{1}{2}x\surd a^2-x^2+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}+C$

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

Calculations :

Given that,

x2 + y2 = 8            -----(1)

⇒ y = √(8 - x²)      -----(2)

Also, the given equation of a line is y = x. Hence, from equation (1)

⇒ x² + x² = 8

⇒ x = ± 2

Now using equation (2), we get y = 2

Hence, we get the intersection of line and circle at (2, 2)

Since we have parabola x2 = 2y. Again using the equation (1)

2y + y² = 8 

⇒ y² + 2y - 8 = 0

⇒ (y + 4)(y - 2) = 0

⇒ y = 2 & -4

Put y = 2 in parabola x² = 2y

⇒ x² = 2 × 2 = 4

⇒ x = ± 2

Hence, we got the intersection of circle and parabola at (-2, 2) and (2, 2).

Required area = area of circle - area of parabola - area of line

⇒ ${\int }_{-2}^2\surd 8-x^2-{\int }_{-2}^0\frac{1}{2}{x}^2-{\int }_0^{2}xdx$

⇒ $2{\int }_0^2\surd 8-x^{2}dx-[\frac{x^3}{6}{]}_{-2}^0-[\frac{x^2}{2}{]}_0^2$

By using the above formula, 

⇒ $2[\frac{x}{2}\surd 8-x^2+4si{n}^{-1}\frac{x}{2\surd 2}]-\frac{4}{3}-2$

⇒ $2[2+4\frac{\pi }{2}]-\frac{10}{3}$

∴ The area bounded is $\frac{2}{3}+2\pi squnits$ and point of intersection is (2, 2).

Concept used:

The area between the curves y1 = f(x) and y2 = g(x) is given by:

Area enclosed = ${\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}|{\mathrm{y}}_2-{\mathrm{y}}_1|\mathrm{d}\mathrm{x}$

Where x1 and x2 are the intersections of curves y1 and y2

Calculation:

In figure ΔABC and ΔAOD is similar

So, Area of the region = 2 × (Area of ΔABC)

For the area of ΔABC

= $\underset{2}{\overset{4}{\int }}\left(\mathrm{x}-2\right)\mathrm{d}\mathrm{x}$

= $[\frac{{\mathrm{x}}^2}{2}-2\mathrm{x}{]}_2^4$     

= $\frac{4^2}{2}-2(4)-(\frac{2^2}{2}-2\times 2)$

= 8 - 8 - 2 + 4

= 2 sq. unit

So, Area of the region = 2 × (Area of ΔABC) = 2 × 2 = 4 sq. unit

Alternate Method

Using the triange formula = 1/2 × base × height

in the figure = 1/2 × (2 × 2)

Area of ΔABC = 2

Total bounded area = 2 × 2 = 4 sq. unit 

Concept:

$\int \sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}+\frac{{\mathrm{a}}^2}{2}\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{\mathrm{x}}{\mathrm{a}}$

Calculations:

Given, the curve is  $\mathrm{x}=\sqrt{9-{\mathrm{y}}^2}$

Now, equation of Y-axis x = 0, 

⇒ y² = 9 

⇒ y = -3, 3

Area bounded by  the curve y = $\sqrt{9-{\mathrm{x}}^2}$ and y-axis 

$=2{\int }_0^3\mathrm{y}\mathrm{d}\mathrm{x}$

$=2{\int }_0^3\sqrt{9-{\mathrm{x}}^2}\mathrm{d}\mathrm{x}$

$=2[\frac{\mathrm{x}}{2}\sqrt{9-{\mathrm{x}}^2}+\frac{9}{2}\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{\mathrm{x}}{3}{]}_0^3$

$=2[\frac{\mathrm{x}}{2}\sqrt{9-{\mathrm{x}}^2}+\frac{9}{2}\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{\mathrm{x}}{3}{]}_0^3$

= 2[$\frac{9\mathrm{\pi }}{4}-0$]

= 4.5 π sq. units 

Hence, option 4 is correct.

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.