# Integral Calculator

This is a calculator which computes the integrals (antiderivative) of a function with respect to a variable x.

Result
\begin{aligned} \displaystyle\int \sin\left(x\right)\, \mathrm d x \\\end{aligned}
\begin{aligned} = -\cos x\ \\\end{aligned}
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### Help

The calculator will computes an integral of given function. To solve definite integrals you need to provide values for upper and lower bound.
1. For powers use ^. Example: x12= x^12 ; ex+2= e^(x+2)
2. For square root use "sqrt". Example: x+1−−−−−√= sqrt(x+1).
3. Supported constants: e, pi
4. Supported functions: sqrt, ln ( use 'ln' instead of 'log'), e (use 'e' instead of 'exp')
5. Please keep 'x' always in '( )'. Example: sin(x)
Trigonometric functions: sin cos tan cot sec csc
Inverse trigonometric functions: acos asin atan acot asec acsc
Hyperbolic functions: sin(h), cosh, tanh, coth, sech, csch

example 1:
$$\int x^2 + 3x - 1 \,dx$$
example 2:
$$\int x^2\,sin x \,dx$$
example 3:
$$\int x^2 + 1 \,dx$$
example 4:
$$\int x \cdot ln x \,dx$$

### Examples of valid and invalid expressions

 Function to integrate Correct syntax is Incorrect syntax is $$(2x+1)^6$$ (2x+1)^6 [2x+1]^6 $$\frac{10x + 1}{x^2-4}$$ (10x+1)/(x^2-4) 10x+1/x^2-4 $$\left(ln(x)\right)^2$$ ln(x)^2 ln^2(x) $$x ~ ln\left(\frac{x-1}{x+1}\right)$$ x*ln((x-1)/(x+1)) x*ln(x-1)/(x+1)

## Introduction to Integral Calculator

Integration can be the inverse of differentiation. The subtraction can be thought the same way that undoing addition, integration undoes differentiation. This type of integration of as called indefinite integration. When you are performing definite integration you integrate to find an area (or region) under a graph, which results in a value for the area. The integral value you work out an and is written is called like this: ⨜fx dx

Calculate the derivative and the limit here – Derivative calculator & Limit Calculator

## Integral calculator with steps

Rules

1. Integration  ⨜ which tells you are integrating.
2. The function which you integrating, f(x). This finding the area under is also the graph you are in the 3 f x = x  This is known as the integrand.
3. The a and b limits. the beginning and end of the region values of a and b define which you are, moving from left-to-right trying to find the area of. b is known as the upper limit and a is known as the lower limit in mathematics. definite integral because it is defined A definite integral is called a limits a and b between the two. They below the integral as like this ⨜ a b are written above and. B=5 and a=2 are the models.
4. The dx at the end. In integration dx dx in differentiation the same meaning as – an amount of x infinitely small. It defines the variable which you also serve to are with respect to integrating. For in this example dx with respect to the variable x  indicates integration, with respect to the variable t dt and many variables, are there.

## Online integral calculator Basic Formula

• ∫x n= x n+1 /n+1 + C
• ∫cos x = sin x + C
• ∫sin x = -cos x + C
• ∫sec 2a = tan a + C
• ∫cosec 2x = -cot x + C
• ∫sec a tan a = sec a + C
• ∫cosec x cot x = -cosec x + C
• ∫da/√ 1- a 2 = sin -1 a + C

### Trigonometric Substitution (a > 0)

• √ a2 − x 2 requires x = a sin θ. Then √ a2 − x 2 = a cos θ, where −π/2 6 θ 6 π/2
• √ a2 + x 2 requires x = a tan θ. Then √ a2 + x 2 = a sec θ, where −π/2 < θ < π/2
• √ x 2 − a2 requires x = a sec θ. Then √ x 2 − a2 = ±a tan θ

– If x > a, use √ x 2 − a2 = +a tan θ, where 0 6 θ < π/2

– If x < a, use √ x 2 − a2 = −a tan θ, where π/2 < θ 6 π

Its purpose is simply to make some elementary connections between the mathematical concept of derivative and various instances of rates of change of physical quantities.

Newton’s Law of Cooling: the rate of cooling of a hot body is proportional to the difference between its temperature and that of the surrounding medium.

Proposition.

Let F be a differentiable real valued function in some open interval containing the point a.

If l := limx→a− F 0 (x) and r := limx→a+ F 0 (x) both exist, then F 0 (a) = r = l = limx→a F 0 (x)

## Strategy for Modeling with Integrals

1. What you want to find Approximate as a sum of values of Riemann continuous function lengths multiplied by the interval. a, b If f x and is the function the interval, and your interval into subintervals partition the of length Δx, the form fx ak Δa with ak the approximating sums will have a point in the fth subinterval.
2. Here a b f x dx, Write a integral definite, the limit to express of these of the partitions go to zero sums as the norms.
3. the integral Evaluate or with an antiderivative numerically.

A car moving with initial velocity of Acceleration of 5 mph accelerates at the rate of at 2.4t mph per second for 8 seconds model the Effects.

(a) the car going how fast is the 8 seconds are up when?

(b) did the car travel how far during 8 seconds those?

SOLUTION

• We first model the effect of the acceleration on the car’s velocity.
• Step 1: Approximate the net change in velocity as a Riemann sum.

, velocity change acceleration when constant time acceleration is applied. To apply this rate of change time formula, we into short subintervals of length Δt partition 0, 8.

On each subinterval, the acceleration is nearly constant, so if tk is any point in the k th subinterval, the change in velocity imparted by the acceleration in the subinterval is approximately atk Δt mph.

Consumption Over Time

A natural tool to calculate net change is an integral and just distance and velocity total accumulation of more than quantities. consumption can be used to calculate decay,  growth, and, next example as in the Integrals. to find the cumulative effect of a varying integrate it the rate of change, whenever you want.

Net Change from Data

Not a fully modeled many real applications, function begin with data. example, we are and asked given data on the rate at which to find the total amount pumped. In the next a pump consecutive 5-minute operates in intervals.

MODEL

∫sin2 a cos2 a da

Solution: Using half-angle identities sin2a = 1-cos(2a)/2 and cos2a= 1+cos(2s)/2 , we get

∫sin2x cos2x dx = ∫ [(1-cos(2x)/2)] × [(1+cos(2x)/2)] dx

= ∫ 1/4 (1-cos2x) dx

= 1/4  ∫1dx – 1/4 ∫cos2x dx

= (x/4) – (1/4) ∫ cos2x dx

On the remaining integral, we apply the half angle identity cos2x = 1+xos4x/2 , and obtain:

∫cos2x dx = ∫ 1/2 dx + ∫ cos4x/2 dx = x/2 + 1/8 cos2 4x +C

Hence,

∫ sin2x cos2x = x/4 – 1/4 (x/2 + 1/8 cos2 4x) +c = x/8 – 1/32 cos2 4x +c