Area under the curve MCQ Quiz - Objective Question with Answer for Area under the curve - Download Free PDF

Calculation

Given $f(0)=0$ & $f^{\prime }(1)=0$ & $f(1)=\frac{1}{2}$

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

$f^{-1}(x)$ 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 $A$

 $=AreaOAB=2[AreaOCM+AreaCMNA-AreaONA]$

$A=2[\frac{1}{2}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{2}(\frac{3}{4}+\frac{1}{2})\times \frac{1}{4}-\frac{1}{2}{\int }_0^1(2x-x^2),dx]$

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

$\Rightarrow A=2[\frac{14}{32}-(1-\frac{1}{3})]$
$\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

Concept:

• The question involves finding the area bounded by inequalities.
• The inequalities form a region enclosed by y = 1/x, 5x − 4y − 1 = 0, 4x + 4y − 17 = 0, and the x-axis.
• To find the area, we:
  • Find the points of intersection of the given curves and lines.
  • Break down the area into simpler regions: triangles and integrals.
  • Use integration to find the area under the curve y = 1/x.
• Integration of 1/x: The integral of 1/x with respect to x is log_ex.
• The final area will be a combination of calculated triangular areas and definite integrals.

Calculation:

Given,

x > 0, y > 1/x, 5x − 4y − 1 > 0, 4x + 4y − 17 < 0

Points of intersection are calculated as follows:

⇒ 5x − 4y − 1 = 0 and y = 1/x meet at (1, 1)

⇒ 5x − 4y − 1 = 0 and 4x + 4y − 17 = 0 meet at (2, 1.25)

⇒ 4x + 4y − 17 = 0 and y = 1/x meet at (4, 0.25)

Break the area into:

Area of triangle with vertices (1,1), (2,1.25), (4,0.25)

Area of region under y = 1/x between x = 1 and x = 4

Area = (1/2) × (base 1.5) × (height 4/3)

⇒ 1/2 × 3/2 × 4/3 = 1

Next, area of another triangle:

Area = (1/2) × (base 2) × (height 10/4)

⇒ 1/2 × 2 × 2.5 = 2.5

Now subtract the area under the curve y = 1/x:

∫₁⁴ (1/x) dx = log_e4

Add all the areas:

Total Area = 1 + 2.5 − log_e4

Total Area = 33/8 − log_e4

∴ Hence, the area of the given region is 33/8 − log_e4.

So, the correct option is 2.

Calculation: 

Abscissa of the point of intersection of 2y = 5x² 

and y = x² + 6 is ± 2 

$\text{ Area }=2{\int }_0^2({\mathrm{x}}^2+6-\frac{5{\mathrm{x}}^2}{2})\mathrm{d}\mathrm{x}={\int }_0^{\frac{2\mathrm{a}}{5}}(\alpha \mathrm{x}-\frac{5{\mathrm{x}}^2}{2})\mathrm{d}\mathrm{x}$

⇒ ${\int }_0^{\frac{2a}{5}}(\alpha x-\frac{5x^2}{2})dx=16$

⇒ α³ = 600

Hence, the correct answer is 600. 

Concept:

Area of Region:

• The area between curves can be found by integrating the difference of the functions over the given interval.
• Use definite integrals and apply limits appropriately to find the enclosed area.

Calculation:

Given region: $\{(x,y):|4-x^2|\le y\le x^2,y\le 4,x\ge 0\}$

Area calculation involves three integrals:

$A={\int }_0^4\sqrt{4+y}dy-{\int }_0^2\sqrt{4-y}dy-{\int }_2^4\sqrt{y}dy$

Evaluating each integral:

${\int }_0^4\sqrt{4+y}dy={[\frac{2}{3}(4+y{)}^{3/2}]}_0^4=\frac{2}{3}(8^{3/2}-4^{3/2})$

${\int }_0^2\sqrt{4-y}dy={[\frac{2}{3}(4-y{)}^{3/2}]}_0^2=\frac{2}{3}(2^{3/2}-4^{3/2})$

${\int }_2^4\sqrt{y}dy={\left[\frac{2}{3}{y}^{3/2}\right]}_2^4=\frac{2}{3}(4^{3/2}-2^{3/2})$

Substitute the values:

$A=\frac{2}{3}(16\sqrt{2}-8)+\frac{2}{3}(2\sqrt{2}-8)-\frac{2}{3}(8-2\sqrt{2})$

$=\frac{40\sqrt{2}-48}{3}=\frac{80\sqrt{2}}{6}-16$

Given: $\frac{80\sqrt{2}}{\alpha }-\beta =\frac{80\sqrt{2}}{6}-16$

Comparing terms,

$\alpha =6,\beta =16$

$\alpha +\beta =6+16=22$

Hence, the correct answer is 22.

Concept:

The area under the curve y = f(x) between x = a and x = b,is given by,  Area = ${\int }_{\mathrm{x}=\mathrm{a}}^{\mathrm{x}=\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}$

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

Calculation:

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

So, the area enclosed by the given curves = ${\int }_0^2\frac{{\mathrm{x}}^2}{2}\mathrm{d}\mathrm{x}$

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

$={\int }_0^2\frac{{\mathrm{x}}^2}{2}\mathrm{d}\mathrm{x}={\left[\frac{{\mathrm{x}}^3}{6}\right]}_0^2$

$=\frac{1}{6}\left(8-0\right)=\frac{4}{3}\mathrm{s}\mathrm{q}.\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$

Hence, option 4 is the correct answer.

Concept:

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

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

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

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

Calculation:

Here, 

x² = y  and line y = 1 cut the parabola

∴ x² = 1

⇒ x = 1 and -1

$\text{Area =}{\int }_{-1}^{1}ydx$

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,

${\text{Area}}_1={\int }_0^{1}ydx$

${\text{Area}}_1={\int }_0^1{x}^{2}dx$

$={\left[\frac{{\mathrm{x}}^3}{3}\right]}_0^1=\frac{1}{3}$

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.

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

$TotalArea=2\times \frac{2}{3}=\frac{4}{3}$ square units.

Concept: 

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

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

Calculation:

Given: 

y = $\surd 16-{\mathrm{x}}^2$ and x-axis

At x-axis, y will be zero

y = $\surd 16-{\mathrm{x}}^2$

⇒ 0 = $\surd 16-{\mathrm{x}}^2$

⇒ 16 - x² = 0

⇒ x2 = 16

∴ x = ± 4

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

Since the curve is y = $\surd 16-{\mathrm{x}}^2$

So, y ≥ o [always]

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

Area of the curve, A $={\int }_{-4}^4\surd 16-{\mathrm{x}}^2\mathrm{d}\mathrm{x}$

We know that,

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

= $[\frac{\mathrm{x}}{2}\surd (4^2-{\mathrm{x}}^2)+\frac{16}{2}\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{\mathrm{x}}{4}{]}_{-4}^4$ 

= $[\frac{\mathrm{x}}{2}\surd (4^2-4^2)+\frac{16}{2}\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{4}{4}]-[\frac{\mathrm{x}}{2}\surd (4^2-(-4{)}^2)+\frac{16}{2}\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{4}{-4})]$

= 8 sin^-1 (1) + 8 sin-1 (1)

= 16 sin^-1 (1)

= 16 × π/2

= 8π sq units

Calculation:

Enclosed Area

$=2{\int }_0^{\pi /4}\left(\cos x-\sin x\right)dx$

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

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

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

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

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 $={\int }_{-2}^2\mathrm{x}\mathrm{d}\mathrm{y}$

$={\int }_{-2}^2(4-{\mathrm{y}}^2)\mathrm{d}\mathrm{y}$

$={[4\mathrm{y}-\frac{{\mathrm{y}}^3}{3}]}_{-2}^2$

$=\frac{32}{3}$ Sq. unit

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

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

Calculation:

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

then 3x² = x² + 4

⇒ x² = 2

⇒ x = ± √ 2

Then the area is 

$\mathrm{A}={\int }_{-\sqrt{2}}^{\sqrt{2}}({\mathrm{x}}^2+4-3{\mathrm{x}}^2)\mathrm{d}\mathrm{x}$

$\mathrm{A}={\int }_{-\sqrt{2}}^{\sqrt{2}}(4-2{\mathrm{x}}^2)\mathrm{d}\mathrm{x}$

$\mathrm{A}=[4\mathrm{x}-2\frac{{\mathrm{x}}^3}{3}{]}_{-\sqrt{2}}^{\sqrt{2}}$

$\mathrm{A}=4[\sqrt{2}-(-\sqrt{2})]-\frac{2}{3}[{\sqrt{2}}^3-{(-\sqrt{2})}^3]$

$\mathrm{A}=8\sqrt{2}-\frac{8}{3}\sqrt{2}$

$\mathrm{A}=\frac{16\sqrt{2}}{3}$ sq unit.

Hence option (4) is correct.

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 = $\underset{0}{\overset{a}{\int }}ydx$ = $\underset{0}{\overset{a}{\int }}\sqrt{a^2-x^2}dx$

Area of circle = 4 × $\underset{0}{\overset{a}{\int }}\sqrt{a^2-x^2}dx$

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 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) = $\left|{\int }_{-2}^{3}4{\mathrm{x}}^3\mathrm{d}\mathrm{x}\right|$

⇒ A = $\left|{\int }_{-2}^{0}4{\mathrm{x}}^3\mathrm{d}\mathrm{x}\right|+\left|{\int }_0^{3}4{\mathrm{x}}^3\mathrm{d}\mathrm{x}\right|$

⇒ A = $\left|4{\left[\frac{{\mathrm{x}}^4}{4}\right]}_{-2}^0\right|+\left|4{\left[\frac{{\mathrm{x}}^4}{4}\right]}_0^3\right|$

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

⇒ A = $|-16|+\left|81\right|$

⇒ A = 97

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 

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,

$RequiredArea={\int }_{-4}^4(16-x^2)dx$

By using Integral property we have,

 $A=2{\int }_0^4(16-x^2)dx$

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

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

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

$=2\times 64\times \frac{2}{3}$

 $A=\frac{256}{3}sq.units$

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:

$Area=2{\int }_0^{16}xdy$

$Area=2{\int }_0^{16}\sqrt{y}dy$

$Area=2\times \frac{2}{3}\times [y^{\frac{3}{2}}{]}_0^{16}$

$Area=2\times \frac{2}{3}\times [{16}^{\frac{3}{2}}-0]$

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

Calculation: 

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

So, the area enclosed by the given curves is given by ${\int }_{-1}^2{\mathrm{x}}^2\mathrm{d}\mathrm{x}$

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 = ${\int }_{-1}^2{\mathrm{x}}^2\mathrm{d}\mathrm{x}$

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

= $[\frac{8}{3}-\frac{-1}{3}]=\frac{9}{3}=3$

Area = 3 sq. units.

Concept:

The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral 

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

This is for curves that are entirely on the same side of the x-axis in the given range.

If the curves are on both sides of the x-axis, then we calculate the areas of both sides separately and add them.

Definite integral: If ∫ f(x) dx = g(x) + C, then ${\displaystyle {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}=[\mathrm{g}(\mathrm{x}){]}_{\mathrm{a}}^{\mathrm{b}}=\mathrm{g}(\mathrm{b})-\mathrm{g}(\mathrm{a}).}$

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

Calculation:

${\displaystyle \int {\mathrm{x}}^4\mathrm{d}\mathrm{x}={\displaystyle \frac{{\mathrm{x}}^5}{5}}+\mathrm{C}}$.

Using the above concept for area of a curve, we can say that the required area is:

$\mathrm{I}={\displaystyle {\int }_1^5{\mathrm{x}}^4\mathrm{d}\mathrm{x}}$

$={\left[{\displaystyle \frac{{\mathrm{x}}^5}{5}}\right]}_1^5$

$={\displaystyle \frac{5^5}{5}}-{\displaystyle \frac{1^5}{5}}$

$={\displaystyle \frac{3125-1}{5}}$

$={\displaystyle \frac{3124}{5}}$.