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

Calculation

Given \(f(0) = 0 \) & \(f'(1) = 0\) & \(f(1) = \frac{1}{2}\)

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

\(f^{-1}(x)\) 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 \(A\)

 \(= AreaOAB = 2[Area OCM + Area CMNA - Area ONA]\)

\(A = 2[\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} + \frac{1}{2} (\frac{3}{4} + \frac{1}{2}) \times \frac{1}{4} - \frac{1}{2} \int_0^1 (2x - x^2) , dx]\)

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

\(\Rightarrow A = 2[\frac{14}{32} -(1 - \frac{1}{3})] \)
\(​\Rightarrow A = \frac{28}{32} - \frac{2}{3} = \frac{7}{8} - \frac{2}{3} \)\(\Rightarrow A = \frac{21 - 16}{24} = \frac{5}{24} \)

⇒ 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 

\(\rm \text { Area }=2 \int_{0}^{2}\left(x^{2}+6-\frac{5 x^{2}}{2}\right) d x=\int_{0}^{\frac{2 a}{5}}\left(α x-\frac{5 x^{2}}{2}\right) d x\)

⇒ \(\int_{0}^{\frac{2 a}{5}}\left(α x-\frac{5 x^{2}}{2}\right) d x=16\)

⇒ 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: \( \{(x,y): |4 - x^2| \leq y \leq x^2, y \leq 4, x \geq 0\} \)

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:

\( \int_0^4 \sqrt{4 + y} \, dy = \left[ \frac{2}{3} (4 + y)^{3/2} \right]_0^4 = \frac{2}{3} \left( 8^{3/2} - 4^{3/2} \right) \)

\( \int_0^2 \sqrt{4 - y} \, dy = \left[ \frac{2}{3} (4 - y)^{3/2} \right]_0^2 = \frac{2}{3} \left( 2^{3/2} - 4^{3/2} \right) \)

\( \int_2^4 \sqrt{y} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_2^4 = \frac{2}{3} \left( 4^{3/2} - 2^{3/2} \right) \)

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,

\( \alpha = 6, \quad \beta = 16 \)

\( \alpha + \beta = 6 + 16 = 22 \)

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 = \(\rm \int_{x=a}^{x =b}f(x) \;dx\)

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

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.

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

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.

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 × \(\mathop \smallint \limits_0^a \sqrt {{a^2} - {x^2}} \;dx\)

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.

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 

​\(\rm \displaystyle \left|\int_a^b f(x)\ dx\right|\)

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 \(\rm \displaystyle \int_a^b f(x)\ dx = [ g(x)]_a^b=g(b)-g(a).\)

\(\rm \displaystyle \int x^n\ dx = \dfrac{x^{n+1}}{n+1}+C\).

Calculation:

\(\rm \displaystyle \int x^4\ dx = \dfrac{x^5}{5}+C\).

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

\(\rm I=\displaystyle \int_1^5 x^4\ dx\)

\(\rm = \left [\dfrac{x^5}{5}\right ]_1^5\)

\(\rm =\dfrac{5^5}{5}-\dfrac{1^5}{5}\)

\(\rm =\dfrac{3125-1}{5}\)

\(\rm =\dfrac{3124}{5}\).