Mathematics Test - 19

Concept:

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

A = \(\rm \displaystyle\int_a^b f(x) \ dx\)

Calculation:

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

∴Area = \(\rm \int _a^bf(y)dy \cdots (\text{function is f(y)})\)

= \(\rm \displaystyle\int_a^b x \ dy\)                   

(∵ f(y) = x)

Hence, option (3) is correct. 

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:

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

Formula used:

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

\(A = \int_a^bf(x)dx\)

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 = \(\rm \int_0^3 [(2x - x^2) - (-x)] dx\)

⇒ A \(= \rm \int_0^3 (3x - x^2) dx = \left[ \frac{3x^2}{2 } - \frac{x^3}{3} \right]_0^3\)

⇒ 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 = \( \rm \int_{x_1}^{x_2}(y_1-y_2)dx\)

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

Calculation:

The 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

Here, the enclosed region is a rectangle as the opposite lines are parallel.

From the graph,

Length of the smaller side = √[(5-4)²+(-1+2)2] = √2 

Length of the larger side = √[(2- (-2))2+(0- 4)2] = 4√2

⇒ A =√2 × 4√2 

⇒ A = 8 sq. units

Alternate Method

Calculation:

The 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

Here, the enclosed region is a rectangle as the opposite lines are parallel.

From the graph,

Length of the smaller side = √[(5-4)2+(-1+2)2] = √2 

Length of the larger side = √[(2- (-2))2+(0- 4)2] = 4√2

⇒ A =√2 × 4√2 

⇒ A = 8 sq. units

Given:

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

Formula used:

\(\int{x^n}.dx = \dfrac{x^{n+1}}{n + 1};\) 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:

\(\int_1^4{x^{1/2}.dx}\)

⇒ \(|\dfrac{x^{3/2}}{3/2}|\), 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 : 

\(\int √{a^2-x^2}dx=\frac{1}{2}x√{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}+C\)

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

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

⇒ \(\int_{-2}^{2} √{8-x^2} - \int_{-2}^{0} \frac{1}{2}x^2- \int_{0}^{2}xdx\)

⇒ \(2\int_{0}^{2}√{8-x^2}dx - [\frac{x^3}{6}]_{-2}^{0} - [\frac{x^2}{2}]_{0}^{2}\)

By using the above formula, 

⇒ \(2[\frac{x}{2}√{8-x^2}+4sin^{-1}\frac{x}{2√{2}}] - \frac{4}{3}-2\)

⇒ \(2[2 + 4\frac{\pi}{2}]- \frac{10}{3}\)

∴ The area bounded is \(\frac{2}{3}+ 2\pi \ sq \ units\) 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 = \(\rm \int_{x_1}^{x_2} |y_2 - y_1|dx\)

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:

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

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:

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 \(\rm \left|\int_a^b f(x)\ dx\right|\), 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.

Calculation:

The graph of the curve y = 2a sin x is presented 

∴ The required area = \(\rm \left|\int_0^{\pi} 2a\sin x\ dx\right|\) = 2a \(\rm \left[-\cos x\right]_0^{\pi}\) =  4a.