Area between two curves MCQ Quiz - Objective Question with Answer for Area between two curves - Download Free PDF

Explanation -

For point of intersection,

4x2 = 8x + 12

⇒ x2 − 2x − 3 = 0

⇒ x = 3, −1

Area bounded is given by
\(A = \int_{-1}^{3} (8x + 12 - 4x^2) \, dx \)

\(A = \left[ \frac{8x^2}{2} + 12x - \frac{4x^3}{3} \right]_{-1}^{3} \)

= 44 – (4/3)

= 128/3

Hence the Correct option is (2)

Calculation

y(x + 4) = 16 ____(1) ,

x + y = 6___(2) 

On solving, (1) & (2)

We get x = 4, x = –2 

Area = \(\int_{-2}^4\left((6-x)-\left(\frac{16}{x+4}\right)\right) d x\)

Area = 30 - 32 Iog_e 2

Calculation

Required area A = \(\int_{1-e}^{0} y \mathrm{dx}=\int_{1-e}^{0} \log _{e}(x+e) \mathrm{dx}\) put x + e = t ⇒ dx = dt also when x = 1 - e, t = 1 and when x = 0, t = e

∴ \(A=\int_{1}^{e} \log _{e} t \mathrm{dt}=\left[t \log _{e} t-t\right]_{1}^{e}\) 

e - e - 0 + 1 = 1

Concept:

Area under the Curve:

• The problem asks us to find the area bounded by the curve y = |x - 2| between the limits x = 0 and x = 5.
• The function y = |x - 2| represents a V-shaped curve, which has a vertex at x = 2. We need to compute the area under this curve between x = 0 and x = 5.
• The function can be split into two parts depending on whether x is less than 2 or greater than 2:
  • For x < 2, y = 2 - x
  • For x ≥ 2, y = x - 2
• To compute the area, we need to split the integral into two parts: one from x = 0 to x = 2, and the other from x = 2 to x = 5.

Calculation:

We are given the function:

y = |x - 2|

Step 1: Split the integral based on the behavior of the function:

I = ∫₀² (2 - x) dx + ∫₂⁵ (x - 2) dx

Step 2: Compute the first integral from x = 0 to x = 2:

∫₀² (2 - x) dx = [2x - (x²/2)]₀²

At x = 2:

2(2) - (2²/2) = 4 - 2 = 2

At x = 0:

2(0) - (0²/2) = 0

So the value of the first integral is:

2 - 0 = 2

Step 3: Compute the second integral from x = 2 to x = 5:

∫₂⁵ (x - 2) dx = [(x²/2) - 2x]₂⁵

At x = 5:

(5²/2) - 2(5) = (25/2) - 10 = 12.5 - 10 = 2.5

At x = 2:

(2²/2) - 2(2) = (4/2) - 4 = 2 - 4 = -2

So the value of the second integral is:

2.5 - (-2) = 2.5 + 2 = 4.5

Step 4: Add the results of the two integrals:

I = 2 + 4.5 = 6.5

∴ The area under the curve is 6.5 square units.

The correct answer is Option (3) 

Concept:

Area between two curves:

Area = ∫ [f(x) - g(x)] dx

• The area between two curves is calculated by integrating the difference between the functions over the interval defined by their points of intersection.
• The general formula for the area between two curves y = f(x) and y = g(x) is:
• In this case, the curves given are y = x and y = x³. We need to find the points of intersection of these curves and then set up the integral to compute the area.

Calculation:

We are given the curves y = x and y = x³. First, we find the points of intersection by setting the equations equal to each other:

x = x³

⇒ x³ - x = 0

⇒ x(x² - 1) = 0

⇒ x(x - 1)(x + 1) = 0

Thus, the points of intersection are x = -1, 0, and 1.

Now, the area between the curves is given by:

Area = ∫ from -1 to 1 (x - x³) dx

We can split the integral into two parts:

∫ (x - x³) dx = ∫ x dx - ∫ x³ dx

Now, integrate each part:

∫ x dx = (x² / 2)

∫ x³ dx = (x⁴ / 4)

So, the integral becomes:

Area = [(x² / 2) - (x⁴ / 4)] evaluated from -1 to 1.

Now, we evaluate the integral at the points x = 1 and x = -1:

At x = 1: (1² / 2) - (1⁴ / 4) = 1/2 - 1/4 = 1/4

At x = -1: ((-1)² / 2) - ((-1)⁴ / 4) = 1/2 - 1/4 = 1/4

Thus, the total area is:

Area = (1/4) - (1/4) = 1/2

∴ The area of the region bounded by the curves y = x and y = x³ is 1/2 square units.

Explanation:

Given curves are y = x - 1 and y2 = 2x + 6 

On solving, we get, 

y² = 2(y + 1) + 6

⇒ y² - 2y - 8 = 0

⇒ (y - 4)(y + 2) = 0

⇒ y = -2, 4

Now, we can find the area by

A = \(\int_{-2}^{4}\left [y+1-\left ( \frac{y^{2}}{2}-3 \right ) \right ]dy \)

= \(\int_{-2}^{4}\left (4+y-\frac{y^{2}}{2} \right )dy \)

= \(\left [ 4y+\frac{y^{2}}{2} -\frac{y^{3}}{6}\right ]_{-2}^{4}\)

= \(16+8-\frac{32}{3}-\left ( -8+2+\frac{4}{3} \right )\)

∴ A = 18

Concept:

1. Standard Equation of Parabola

(y - y₁)² = 4a(x - x₁)

Where (x₁,y₁) is vertex and focus is (x₁ + a,y₁)

Calculation:

y² = 6 (x – 1) and y² = 3x

From the diagram

Area bounded by both parabola is

⇒ A = 2 × \(\left[ {\mathop \smallint \nolimits_0^{\sqrt 6 } \left( {\frac{{{{\rm{y}}^2}}}{6} + 1} \right){\rm{dy\;}} - {\rm{\;\;}}\mathop \smallint \nolimits_0^{\sqrt 6 } \frac{{{{\rm{y}}^2}}}{3}{\rm{\;dy}}} \right]\)

⇒ A = 2 × \(\left[ {\left[ {\frac{{{{\rm{y}}^3}}}{{18}} + {\rm{y}}} \right]\begin{array}{*{20}{c}} {\sqrt 6 }\\ 0 \end{array} - \left[ {\frac{{{{\rm{y}}^3}}}{9}} \right]\begin{array}{*{20}{c}} {\sqrt 6 }\\ 0 \end{array}} \right]\)

⇒ A = 2 × \(\left[ {\left[ {\frac{{6\sqrt 6 }}{{18}} + \sqrt 6 - 0{\rm{\;}}} \right] - \frac{{6\sqrt 6 }}{9} - 0{\rm{\;}}} \right]\)

⇒ A = 2 × \(\left[ { - {\rm{\;}}\frac{{6\sqrt 6 }}{{18}} + \sqrt 6 {\rm{\;}}} \right]\)

Concept:

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

Area enclosed = \(\rm \left|\int_{x_1}^{x_2}(y_1-y_2)dx\right|\)

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

Calculation:

Given

Curve 1: y = x2 = f(x) (say)

Curve 2: y - x = 0

⇒ y = 8 - 6x = g(x) (say)

To find the intersections (or limits of the area) putting value of y from curve 1

⇒ x² - x = 0

⇒ x(x - 1) = 0

⇒ x = 0, x = 1

Now the required area (A) is 

A = \(\rm \left|\int_{x_1}^{x_2}[f(x)-g(x)]dx\right|\)

⇒ A = \(\rm \left|\int_{0}^{1}[x - x^2]dx\right|\)

⇒ A = \(\rm \left|\left[{x^2\over2}-{x^3\over3}\right]_{0}^{1}\right|\)

⇒ A = \(\left| {\frac{1}{2} - \;\frac{1}{3} - \left( 0 \right)} \right|\)

⇒ A = \(\rm 1\over6\)sq. unit

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 

Concept used:

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

Area enclosed = \(\rm \int_{x_1}^{x_2} |y_2 - y_1|dx\)

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

= \(\rm \mathop \smallint \limits_2^4 \left( {x - 2} \right)\;dx\)

= \(\rm[\frac{x^2} 2 - 2x ]^{4}_2\)     

= \( \frac {4^2}2 - 2(4) - (\frac {2^2}2 - 2\times2)\)

= 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:

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

Concept:

Area under a Curve by Integration: Find the area under this curve is by summing vertically.

In this case, we find the area is the sum of the rectangles, heights 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 bottom to top.

\({\rm{Area}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{xdy}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{y}} \right){\rm{dy}}\)

Calculation:

Given curve is x² = 4y

∴ x = 2√y

Here required shaded area of the region lying in first quadrant bounded by parabola x² = 4y, and the horizontal lines y = 2 and y = 4.

\({\rm{Area\;of\;ABCD}} = \mathop \smallint \nolimits_2^4 {\rm{xdy}}\)

\(= \mathop \smallint \nolimits_2^4 2\sqrt {\rm{y}} {\rm{\;dy}} = 2\left[ {\frac{{{y^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_2^4 = \frac{4}{3}\left( {{4^{\frac{3}{2}}} - {2^{\frac{3}{2}}}} \right) = \frac{{32 - 8\sqrt 2 }}{3}{\rm{sq}}.{\rm{\;unit}}\)

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 required.

Calculation:

The f(x) = y = 2e4x 

Given the end points x₁ = 0, x₂ = 4

Area of the curve (A) = |\(\rm \int_0^4\)2e4x dx|

⇒ A = \(\rm \left|\left[2e^{4x}\over4\right]_0^4\right|\)

⇒ A = \(\rm {1\over2}\left|\left[e^{4x}\right]_0^4\right|\)

⇒ A = \(\rm {1\over2}\left|\left[e^{16}-e^0\right]\right|\)

⇒ A = \(\boldsymbol{\rm {1\over2}\left(e^{16}-1\right)}\)

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 

Concept:

\(\rm \int\sqrt{a^2-x^2}=\frac x 2\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac x a\)

Calculations:

Given, the curve is  \(\rm x = \sqrt {9 - y^2}\)

Now, equation of Y-axis x = 0, 

⇒ y² = 9 

⇒ y = -3, 3

Area bounded by  the curve y = \(\rm \sqrt{9-x^2}\) and y-axis 

\(=\rm 2\int_0^3ydx\)

\(=\rm 2\int_0^3\rm \sqrt{9-x^2}dx\)

\(\rm =2[\frac x2\sqrt{9-x^2}+\frac{9}{2}sin^{-1}\frac x 3]_0^3\)

\(\rm =2[\frac x2\sqrt{9-x^2}+\frac{9}{2}sin^{-1}\frac x 3]_0^3\)

= 2[\(\rm \frac{9π}{4}-0\)]

= 4.5 π sq. units 

Hence, option 4 is correct.

Concept:

• The area under the function y = f(x) from x = a to x = b and the x-axis is given by \(\left|\int_{a}^{b}f(x)dx\right|\), 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.
• If we have curves y = f(x) and y = g(x), then the area between the curves for a < x < b if given by \(\left|\int_{a}^{b}[f(x)-g(x)]dx\right|\).

Calculation: