Exercise 1

Calculate the area of the site bounded by the curve y = 4x - x^2  and the x-axis.

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Exercise 3

Find the area bounded by the line x + y = 10, the x-axis, x = 2 and x = 8.

Exercise 2

Find the area of the plane region enclosed by the curve y = ln x between the point of intersection with the x-axis and x = e.

Exercise 4

Calculate the area enclosed by the curve y = 6x^2 - 3x^3  and the x-axis.

Exercise 5

Calculate the area enclosed by the curve f(x) = x^3 - 6x^2 + 8x and the x-axis.

Exercise 6

Calculate the area of a circle of radius r.

Exercise 7

Find the area of an ellipse of semiaxes a and b.

Exercise 8

 Calculate the area enclosed by the curve y = x^2 - 5x + 6  and the line y = 2x.

Exercise 9

Calculate the area enclosed by the parabola y^2 = 4x  and the line y = x.

Exercise 10

Calculate the area enclosed by 3y = x^2 and y = -x^2 + 4x.

Exercise 11

Calculate the area enclosed by y = x^2 - 2x and y = - x^2 + 4x.

Exercise 12

Calculate the area enclosed by:

y = sin x, y = cos x, x = 0.

 

 

Solution of exercise 1

Calculate the area of the site bounded by the curve y = 4x - x^2 and the x-axis.

First, find the x-intercepts to the curve and the limits of integration.

0 = 4x - x^2

x = 0

x = 4

Graph of the Example 1

A = \int^4_0 (4x - x^2) dx = [2x^2 - \frac{x^3}{3}]^4_0 = \frac{32}{3}u^2

 

Solution of exercise 2

Find the area of the plane region enclosed by the curve y = ln x between the point of intersection with the x-axis and x = e.

Graph of the Example 2

First, find the x-intercepts.

ln x = 0        e^0 = 1

\int^e_1 lnx dx

\int lnx dx = x ln x - \int dx = x ln x - x + C

\int^e_1 lnx dx = [x (lnx - 1)]^e_1 = 0 + 1 = 1 u^2

 

Solution of exercise 3

Find the area bounded by the line x + y = 10, the x-axis, x = 2 and x = 8.

Graph of the Example 3

A = \int^8_2 (10 - x)dx = [10x - \frac{x^2}{2}]^8_2 = 30u^2

 

Solution of exercise 4

Calculate the area enclosed by the curve y = 6x^2 - 3x^3 and the x-axis.

6x^2 - 3x^3 = 0

3x^2 (2 - x) = 0

x_1 = 0       x_2 = 2

Graph of the Example 4

A = \int^2 _ 0(6x^2 - 3x^3) dx = [2x^3 - \frac{3}{4}x^4]^2_0 = 16 - 12 = 4u^2

 

Solution of exercise 5

Calculate the area enclosed by the curve f(x) = x^3 - 6x^2 + 8x and the x-axis.

x^3 - 6x^2 + 8x = 0

x (x^2 - 6x + 8) = 0

x = 0

x = 2

x = 4

Graph of the Example 5

A = \int^2_0 (x^3 - 6x^2 + 8x) dx + |\int^4_2 (x^3 - 6x^2 + 8x)dx|

The area, for reasons of symmetry, can be written as:

A = 2 \int^2 _0 (x^3 - 6x^2 + 8x)dx = 2 [\frac{x^4}{4} - 2x^3 + 4x^2]^2_0 = 8u^2

 

Solution of exercise 6

Calculate the area of a circle of radius r.

 

Start from the equation of the circle x^2 + y^2 = r^2.

Graph of the Example 6

The area of the circle is four times the area of the first quadrant.

A_1 = \int^r_0 \sqrt{r^2 - x^2} dx

Calculate the indefinite integral by change of variable.

\int \sqrt{r^2 - x^2} dx

x = r sin t

dx = r cost t dt

\int^r_0 \sqrt{r^2 - x^2} = \int \sqrt{r^2 - r^2 sin^2t} r cos tdt = \int \sqrt{(r^2 (1 - sin^2t)} r cost dt

= \int r^2 cos^2t dt = r^2 \int cos^2 t dt = \int \frac{1 + cos2t}{2}dt = r^2 [\frac{t}{2} + \frac{1}{4} sen 2t] + C

Find the new limits of integration.

x = 0        0 = r sin t                sin t = 0           t = 0

x = r         r = r sin t                sin t = 1             t = \frac{\pi}{2}

A_1 = r^2 [\frac{t}{2} + \frac{1}{4} sen 2t]^{\frac{\pi}{2}} _0 = r^2 (\frac{\pi}{4} - 0) = \frac{1}{4} \pi r^2

A = 4 A_1 = \pi r^2

 

Solution of exercise 7

Find the area of an ellipse of semiaxes a and b.

Graph of the Example 7

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

y = \pm \frac{b}{a} \sqrt{a^2 - x^2}

As the ellipse is a symmetrical curve, the area requested will be 4 times the area enclosed in the first quadrant of the coordinate axes.

A = 4 \int^a_0 \frac{b}{a} \sqrt{a^2 - x^2} dx

\int \sqrt{a^2 - x^2} dx

x = a sin t

dx = a cos t dt

\int^r_0 = \sqrt{a^2 - x^2} = \int \sqrt{a^2 - a^2 sin^2 t} a cost dt = \int \sqrt{a^2 (1 - sin^2 t)} a cos t dt

= \int a^2 cos^2 t dt = a^2 \int cos^2 t dt = a^2 \int \frac{1 + cos2t}{2} dt = a^2 [ \frac{t}{2} + \frac{1}{4} sin^2 2t] + C

Find the new limits of integration.

x = 0           0 = r sin t        sin t = 0       t = 0

x = a           a = a sin t         sin t = 1       t = \frac{\pi}{2}

A = 4 \int^a_0 \frac{b}{a} \sqrt{a^2 - x^2} dx = \frac{4b}{a} \cdot a^2 [\frac{t}{2} + \frac{1}{4} sen^2 2t]^{\frac{\pi}{2}}_0 = 4ab(\frac{\pi}{4}) = \pi ab

 

Solution of exercise 8

Calculate the area enclosed by the curve y = x^2 - 5x + 6  and the line y = 2x.

First, find the points of intersection of the two functions to know the limits of integration.

y = x^2 - 5x + 6, y = 2x

x_1 = 1

x_2 = 6

Graph of the Example 8

From x = 1 to x = 6, the line is above the parabola.

A = \int^6_1 (2x - x^2 + 5x - 6) dx = \int ^6_1 (-x^2 + 7x - 6)dx = [-\frac{x^3}{3} + \frac{7x^2}{2} - 6x]^6_1

= (-\frac{6^3}{3} + \frac{7 \cdot 6^2}{2} - 36) - (- \frac{1}{3} + \frac{7}{2} - 6) = \frac{125}{6}u^2

 

Solution of exercise 9

Calculate the area enclosed by the parabola y^2 = 4x  and the line y = x.

y^2 = 4xy = x

y^2 = 4y    (0, 0) (4,0)

Graph of the Example 9

From x = 0 to x = 4, the parabola is above the line.

A = \int^4 _ 0 \sqrt{4x} dx - \int ^4_0 x dx = \int ^4_0 (\sqrt{4x} - x) dx = [\frac{4}{3} x^{\frac{3}{2}} - \frac{x^2}{2}]^4_0 = \frac{8}{3} u^2

 

Solution of exercise 10

Calculate the area enclosed by 3y = x^2 and y = - x^2 + 4x.

First, represent the parabolas from the vertex and the points of intersection with the axes.

y = \frac{x^2}{3}

x_y = 0

y_y = 0

V(0,0)

y = - x^2 + 4x

x_y = - \frac{4}{-2} = 2

y_y = 4

V(2, 4)

-x^2 + 4x = 0

x_1 = 0

x_2 = 4

Also, find the points of intersection of the functions, which will give the limits of integration.

y = \frac{x^2}{3}

y = - x^2 + 4x

(0,0)     (3,3)

Graph of the Example 10

A = \int^3_0 (-x^2 + 4x - \frac{x^2}{3}) dx = \int^3_0 (- \frac{4}{3} x^2 + 4x) dx

= [- \frac{4}{3} x^3 + 2x^2]^3_0 = -12 + 18 = 6u^2

 

Solution of exercise 11

Calculate the area enclosed by y = x^2 - 2x and y = -x^2 + 4x.

Represent the parabolas from the vertex and the points of intersection with the axes.

x_y = \frac{2}{2} = 1           y_y = 1^2 - 2 \cdot 1 = -1               V (1, -1)

0 = x^2 - 2x                            0 = x(x - 2)                    (0, 0)       (2, 0)

x_y = \frac{-4}{-2} = 2            y_y = -2^2 + 4 \cdot 2 = 4             V (2, -4)

0 = - x^2 + 4x             0 = x(-x + 4)             (0,0)          (4, 0)

y = x^2 - 2x,    y = -x^2 + 4x

x^2 - 2x = -x^2 + 4x

(0, 0)

(3, 3)

Graph of the Example 11

A_1 = \int^2 _ 0 (x^2 - 2x) dx = [\frac{x^3}{3} - x^2]^2 _0 = - \frac{4}{3}        |A_1| = \frac{4}{3} u^2

A_2 = \int^3_0 (-x^2 + 4x) dx = [- \frac{x^3}{3} + 2x^2]^3_0 = 9     A_2 = 9u^2

A_3 = \int ^3_2(x^2 - 2x) dx = [\frac{x^3}{3} - x^2}]^3_2 = \frac{4}{3}          A_3 = \frac{4}{3}u^2

A = |A_1| + A_2 - A_3

A = \frac{4}{3} + 9 - \frac{4}{3} = 9u^2

 

Solution of exercise 12

Calculate the area enclosed by:

y = sin x, y = cos x, x = 0.

First, find the points of intersection of the functions:

y = sin x, y = cos x

sin x = cos x

x = \frac{\pi}{4}

The cosine graph is above the graph within the limits of integration.

A = \int ^ {\frac{\pi}{4}}_0 (cos x - sin x) dx = [sin x + cos x]^{\frac{\pi}{4}} _ 0 = (\sqrt{2} - 1) u^2

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.