Integral of 1

Before going to find the integral of 1, let us recall how do we integrate very basic functions like 1, x, sin x, etc? Since integration is the inverse process of differentiation, we can just use the differentiation to do the integration as well. i.e., for knowing what is the result of the integration of 1, we have to think by differentiating what function would result in 1.

Let us learn more about the integral of 1 by different ways of finding it. Also, we will solve a few examples on the same.

The integral of 1 with respect to x is x + C. This is mathematically written as ∫ 1 dx = x + C. Here,

• 1 is the integrand.
• dx denotes that the integration is with respect to x.
• C is the constant of integration.

Let us see how to prove the integration of 1 formula in two ways:

• using differentiation
• using power rule of integration

For finding the integral of 1 using the process of differentiation, see by differentiating what expression would give 1. i.e., think

d/dx ( ? ) = 1

We know that the derivative of x is 1. So

d/dx (x) = 1

Taking the integral on both sides

∫ d/dx (x) dx = ∫ 1 dx

By the fundamental theorem of calculus, the integral and derivative get canceled. So we get

x = ∫ 1 dx

We actually add an integration constant for all indefinite integrals. So

∫ 1 dx = x + C.

Hence, we have derived the formula of integration of 1.

Verification of Integral of 1

To verify the integral of 1, we just differentiate the result and see whether we get 1 back. Since the ∫ 1 dx = x + C, let us find the derivative of x + C. Then

d/dx (x + C) = d/dx (x) + d/dx (C) (by derivative rules)

= 1 + 0

= 1

Therefore, the integral of 1 is x + C and is verified.

Let us consider the integral ∫ 1 dx. Using the properties of exponents, 1 = x⁰. Then the integral becomes ∫ x⁰ dx. By using the power rule of integration,

∫ xⁿ dx = x^{n + 1}/(n +1) + C

Substitute n = 0 here,

∫ x⁰ dx = x^{0 + 1}/(0 +1) + C

∫ 1 dx = x¹/1 + C

∫ 1 dx = x + C

Hence, the integral of 1 formula is derived.

The definite integral of 1 is the integral of 1 with the lower and upper limits. Let us consider a definite integral with the lower limit a and upper limit b. i.e., ∫ₐ ^b 1 dx. Since ∫ 1 dx = x + C, the definite integral value is obtained by substituting the upper and lower limit and subtracting the results. Then

∫ₐ ^b 1 dx = (b + C) - (a + C) = b - a.

So the definite integral of 1 is always equal to the difference between the upper limit and lower limit.

Important Notes on Integral of 1:

• The integral of 1 is x + C. i.e., ∫ 1 dx = x + C.
• Hence, the integral of any constant is, ∫ a dx = a ∫ 1 dx = ax + C.
• The definite integral from a to b is b - a. i.e., ∫ₐ ^b 1 dx = b - a.

Related Topics:

• Derivative Calculator
• Derivative Formulas
• Inverse Trig Derivatives
• Differentiation of Trigonometric Functions

What is the value of Integration of 1?

The integration of 1 is x + C. It is written as ∫ 1 dx = x + C, where C is the integration constant.

How to Do the Integration of 1?

To find the integral of 1, just search the derivative formulas and see by differentiating what function would result in 1. We have d/dx (x + C) = 1, where C is any constant. Hence the integral of 1 is x + C.

Is the Antiderivative of 1 Equal to 1 Itself?

No, the antiderivative of 1 is equal to x + C. Another name for the antiderivative is integral and hence the integral of 1 is x + C which is written as ∫ 1 dx = x + C.

What is the Integral of 2 Using the Fact that Integral of 1 is x?

It is given that ∫ 1 dx = x + C. Using this, ∫ 2 dx = 2 ∫ 1 dx = 2x + C.

How to Find the Definite Integral of 1?

We have ∫ 1 dx = x + C. If we consider the definite integral, ∫ₐ ^b 1 dx and to evaluate this, first we should substitute x = b and then x = a in the result x + C and find the difference. Then ∫ₐ ^b 1 dx = (b+C) - (a+C) = b - a.