Enter the function for which you want to find the integral

The calculator provides a DETAILED solution of definite integrals.

This calculator solves the definite integral of the function f(x) with the given upper and lower limits.

Examples

With the use of degree
(square and cube) and fractions

(x^2 - 1)/(x^3 + 1)

Square root

Sqrt(x)/(x + 1)

cube root

Cbrt(x)/(3*x + 2)

Using sine and cosine

2*sin(x)*cos(x)

Arcsine

X*arcsin(x)

Arc cosine

x*arccos(x)

Application of the logarithm

X*log(x, 10)

natural logarithm

Exhibitor

Tg(x)*sin(x)

Cotangent

Ctg(x)*cos(x)

Irrational fractions

(sqrt(x) - 1)/sqrt(x^2 - x - 1)

Arctangent

X*arctg(x)

Arc tangent

X*arсctg(x)

Hyberbolic sine and cosine

2*sh(x)*ch(x)

Hyberbolic tangent and cotangent

ctgh(x)/tgh(x)

Hyberbolic arcsine and arccosine

X^2*arcsinh(x)*arccosh(x)

Hyberbolic arctangent and arccotangent

X^2*arctgh(x)*arctgh(x)

Rules for entering expressions and functions

Expressions can consist of functions (notations are given in alphabetical order): absolute(x) Absolute value x
(module x or |x|) arccos(x) Function - arc cosine of x arccosh(x) Arc cosine hyperbolic from x arcsin(x) Arcsine from x arcsinh(x) Arcsine hyperbolic from x arctg(x) Function - arc tangent from x arctgh(x) The arc tangent is hyperbolic from x e e a number that is approximately equal to 2.7 exp(x) Function - exponent from x(which is e^x) log(x) or log(x) Natural logarithm of x
(To obtain log7(x), you need to enter log(x)/log(7) (or, for example, for log10(x)=log(x)/log(10)) pi The number is "Pi", which is approximately equal to 3.14 sin(x) Function - Sine of x cos(x) Function - Cosine of x sinh(x) Function - Hyperbolic sine of x cash(x) Function - Hyperbolic cosine of x sqrt(x) The function is the square root of x sqr(x) or x^2 Function - Square x tg(x) Function - Tangent from x tgh(x) Function - Hyperbolic tangent of x cbrt(x) The function is the cube root of x

You can use the following operations in expressions: Real numbers enter in the form 7.5 , not 7,5 2*x- multiplication 3/x- division x^3- exponentiation x + 7- addition x - 6- subtraction
Other features: floor(x) Function - rounding x down (example floor(4.5)==4.0) ceiling(x) Function - rounding x up (example ceiling(4.5)==5.0) sign(x) Function - Sign x erf(x) Error function (or probability integral) laplace(x) Laplace function

Calculate the area of ​​a figure bounded by lines.

Solution.

We find the points of intersection of the given lines. To do this, we solve the system of equations:

To find the abscissas of the points of intersection of the given lines, we solve the equation:

We find: x 1 = -2, x 2 = 4.

So, these lines, which are a parabola and a straight line, intersect at points A(-2; 0), B(4; 6).

These lines form a closed figure, the area of ​​\u200b\u200bwhich is calculated using the above formula:

According to the Newton-Leibniz formula, we find:

Find the area of ​​an area bounded by an ellipse.

Solution.

From the ellipse equation for the I quadrant we have . From here, according to the formula, we obtain

Let's apply the substitution x = a sin t, dx = a cos t dt. New limits of integration t = α and t = β are determined from the equations 0 = a sin t, a = a sin t. Can be put α = 0 and β = π /2.

We find one fourth of the required area

From here S = pab.

Find the area of ​​a figure bounded by linesy = - x 2 + x + 4 andy = - x + 1.

Solution.

Find the intersection points of the lines y = -x 2 + x + 4, y = -x+ 1, equating the ordinates of the lines: - x 2 + x + 4 = -x+ 1 or x 2 - 2x- 3 = 0. Find the roots x 1 = -1, x 2 = 3 and their corresponding ordinates y 1 = 2, y 2 = -2.

Using the figure area formula, we get

Find the area enclosed by the parabolay = x 2 + 1 and directx + y = 3.

Solution.

Solving the system of equations

find the abscissas of the intersection points x 1 = -2 and x 2 = 1.

Assuming y 2 = 3 - x and y 1 = x 2 + 1, based on the formula we get

Calculate the area contained within the Bernoulli lemniscater 2 = a 2 cos 2 φ .

Solution.

In the polar coordinate system, the area of ​​the figure bounded by the arc of the curve r = f(φ ) and two polar radii φ 1 = ʅ and φ 2 = ʆ , is expressed by the integral

Due to the symmetry of the curve, we first determine one-fourth of the desired area

Therefore, the total area is S = a 2 .

Calculate the arc length of an astroidx 2/3 + y 2/3 = a 2/3 .

Solution.

We write the equation of the astroid in the form

(x 1/3) 2 + (y 1/3) 2 = (a 1/3) 2 .

Let's put x 1/3 = a 1/3 cos t, y 1/3 = a 1/3 sin t.

From here we obtain the parametric equations of the astroid

x = a cos 3 t, y = a sin 3 t, (*)

where 0 ≤ t ≤ 2π .

In view of the symmetry of the curve (*), it suffices to find one fourth of the arc length L corresponding to the parameter change t from 0 to π /2.

We get

dx = -3a cos 2 t sin t dt, dy = 3a sin 2 t cos t dt.

From here we find

Integrating the resulting expression in the range from 0 to π /2, we get

From here L = 6a.

Find the area bounded by the spiral of Archimedesr = aφ and two radius vectors that correspond to polar anglesφ 1 andφ 2 (φ 1 < φ 2 ).

Solution.

Area bounded by a curve r = f(φ ) is calculated by the formula , where α and β - limits of change of the polar angle.

Thus, we get

(*)

From (*) it follows that the area bounded by the polar axis and the first turn of the Archimedes spiral ( φ 1 = 0; φ 2 = 2π ):

Similarly, we find the area bounded by the polar axis and the second turn of the Archimedes spiral ( φ 1 = 2π ; φ 2 = 4π ):

The required area is equal to the difference of these areas

Calculate the volume of a body obtained by rotation around an axisOx figure bounded by parabolasy = x 2 andx = y 2 .

Solution.

Let's solve the system of equations

and get x 1 = 0, x 2 = 1, y 1 = 0, y 2 = 1, whence the intersection points of the curves O(0; 0), B(eleven). As can be seen in the figure, the desired volume of the body of revolution is equal to the difference between the two volumes formed by rotation around the axis Ox curvilinear trapezoids OCBA and ODBA:

Calculate the area bounded by the axisOx and sinusoidy = sinx on segments: a); b) .

Solution.

a) On the segment, the function sin x preserves the sign, and therefore by the formula , assuming y= sin x, we find

b) On the segment , function sin x changes sign. For the correct solution of the problem, it is necessary to divide the segment into two and [ π , 2π ], in each of which the function retains its sign.

According to the rule of signs, on the segment [ π , 2π ] area is taken with a minus sign.

As a result, the desired area is equal to

Determine the volume of the body bounded by the surface obtained from the rotation of the ellipsearound the major axisa .

Solution.

Given that the ellipse is symmetrical about the coordinate axes, it is enough to find the volume formed by rotation around the axis Ox area OAB, equal to one quarter of the area of ​​the ellipse, and double the result.

Let us denote the volume of the body of revolution through V x; then, based on the formula, we have , where 0 and a- abscissas of points B and A. From the equation of the ellipse we find . From here

Thus, the required volume is equal to . (When the ellipse rotates around the minor axis b, the volume of the body is )

Find the area bounded by parabolasy 2 = 2 px andx 2 = 2 py .

Solution.

First, we find the coordinates of the intersection points of the parabolas in order to determine the integration interval. Transforming the original equations, we obtain and . Equating these values, we get or x 4 - 8p 3 x = 0.

x 4 - 8p 3 x = x(x 3 - 8p 3) = x(x - 2p)(x 2 + 2px + 4p 2) = 0.

We find the roots of the equations:

Considering the fact that the point A the intersection of the parabolas is in the first quarter, then the limits of integration x= 0 and x = 2p.

The desired area is found by the formula

Example1 . Calculate the area of ​​the figure bounded by lines: x + 2y - 4 = 0, y = 0, x = -3, and x = 2

Let's build a figure (see Fig.) We build a straight line x + 2y - 4 \u003d 0 along two points A (4; 0) and B (0; 2). Expressing y in terms of x, we get y \u003d -0.5x + 2. According to formula (1), where f (x) \u003d -0.5x + 2, a \u003d -3, b \u003d 2, we find

S \u003d \u003d [-0.25 \u003d 11.25 sq. units

Example 2 Calculate the area of ​​​​the figure bounded by lines: x - 2y + 4 \u003d 0, x + y - 5 \u003d 0 and y \u003d 0.

Solution. Let's build a figure.

Let's build a straight line x - 2y + 4 = 0: y = 0, x = - 4, A (-4; 0); x = 0, y = 2, B(0; 2).

Let's construct a straight line x + y - 5 = 0: y = 0, x = 5, С(5; 0), x = 0, y = 5, D(0; 5).

Find the point of intersection of the lines by solving the system of equations:

x = 2, y = 3; M(2; 3).

To calculate the required area, we divide the AMC triangle into two triangles AMN and NMC, since when x changes from A to N, the area is limited by a straight line, and when x changes from N to C, it is a straight line

For triangle AMN we have: ; y \u003d 0.5x + 2, i.e. f (x) \u003d 0.5x + 2, a \u003d - 4, b \u003d 2.

For the NMC triangle we have: y = - x + 5, i.e. f(x) = - x + 5, a = 2, b = 5.

Calculating the area of ​​each of the triangles and adding the results, we find:

sq. units

sq. units

9 + 4, 5 = 13.5 sq. units Check: = 0.5AC = 0.5 sq. units

Example 3 Calculate the area of ​​a figure bounded by lines: y = x 2 , y = 0, x = 2, x = 3.

In this case, it is required to calculate the area of ​​a curvilinear trapezoid bounded by a parabola y = x 2 , straight lines x \u003d 2 and x \u003d 3 and the Ox axis (see Fig.) According to formula (1), we find the area of ​​\u200b\u200ba curvilinear trapezoid

= = 6kv. units

Example 4 Calculate the area of ​​​​a figure bounded by lines: y \u003d - x 2 + 4 and y = 0

Let's build a figure. The desired area is enclosed between the parabola y \u003d - x 2 + 4 and axis Oh.

Find the points of intersection of the parabola with the x-axis. Assuming y \u003d 0, we find x \u003d Since this figure is symmetrical about the Oy axis, we calculate the area of ​​\u200b\u200bthe figure located to the right of the Oy axis, and double the result: \u003d + 4x] square. units 2 = 2 sq. units

Example 5 Calculate the area of ​​a figure bounded by lines: y 2 = x, yx = 1, x = 4

Here it is required to calculate the area of ​​the curvilinear trapezoid bounded by the upper branch of the parabola y 2 \u003d x, the Ox axis and straight lines x \u003d 1x \u003d 4 (see Fig.)

According to formula (1), where f(x) = a = 1 and b = 4, we have = (= sq. units

Example 6 . Calculate the area of ​​the figure bounded by lines: y = sinx, y = 0, x = 0, x= .

The desired area is limited by a half-wave sinusoid and the Ox axis (see Fig.).

We have - cosx \u003d - cos \u003d 1 + 1 \u003d 2 square meters. units

Example 7 Calculate the area of ​​the figure bounded by lines: y \u003d - 6x, y \u003d 0 and x \u003d 4.

The figure is located under the Ox axis (see Fig.).

Therefore, its area is found by the formula (3)

= =

Example 8 Calculate the area of ​​\u200b\u200bthe figure bounded by the lines: y \u003d and x \u003d 2. We will build the curve y \u003d by points (see figure). Thus, the area of ​​\u200b\u200bthe figure is found by the formula (4)

Example 9 .

X 2 + y 2 = r 2 .

Here you need to calculate the area bounded by the circle x 2 + y 2 = r 2 , i.e. the area of ​​a circle of radius r centered at the origin. Let's find the fourth part of this area, taking the limits of integration from 0

dor; we have: 1 = = [

Consequently, 1 =

Example 10 Calculate the area of ​​\u200b\u200bthe figure bounded by lines: y \u003d x 2 and y = 2x

This figure is limited by the parabola y \u003d x 2 and straight line y \u003d 2x (see Fig.) To determine the intersection points of the given lines, we solve the system of equations: x 2 – 2x = 0 x = 0 and x = 2

Using formula (5) to find the area, we obtain

= }