Integral of a Constant

The indefinite integral of a constant \(k\) is found by multiplying the constant by the variable of integration and adding an arbitrary constant of integration \(C\). $$ \int k\,dx=kx+C $$

Why does this formula work? Because the derivative of \(kx+C\) with respect to \(x\) is exactly \(k\).

$$ \frac{d}{dx}(kx+C)=k $$

For example, suppose I want to find the indefinite integral of the constant function \(f(x)=3\).

$$ \int 3\,dx $$

Applying the formula, I get

$$ \int 3\,dx=3x+C $$

Geometrically, this result represents a family of parallel straight lines, all with a slope of 3. The constant of integration \(C\) determines where each line crosses the y-axis.

Definite Integral of a Constant

The definite integral of a constant \(k\) over an interval \([a,b]\) is calculated using the formula $$ \int_a^b k\,dx=k(b-a) $$

To understand where this formula comes from, consider the case where \(k>0\) and \(a< b\).

The graph of the constant function \(f(x)=k\) is a horizontal line parallel to the x-axis. Between \(x=a\) and \(x=b\), the region below the graph and above the x-axis forms a rectangle.

This rectangle has a width of \(b-a\) and a height of \(k\). Since the area of a rectangle is its width multiplied by its height, the definite integral is simply

$$ \int_a^b k\,dx=k(b-a) $$

For example, consider the integral of the constant function \(f(x)=3\) from \(x=1\) to \(x=4\).

$$ \int_1^4 3\,dx=3(4-1)=9 $$

The result is 9, which is the area of a rectangle with a width of 3 and a height of 3.

What happens if the constant is negative?

When \(k<0\), the graph lies below the x-axis. The definite integral is then negative because it represents a signed area.

The geometric area of the rectangle, however, is always nonnegative. To calculate it, I use the absolute value of the constant.

$$ A=|k|(b-a) $$

If \(k=0\), the graph coincides with the x-axis, and the definite integral is zero.

Note. The indefinite integral of a constant gives a family of functions of the form \(F(x)=kx+C\), represented by parallel straight lines. The definite integral, on the other hand, gives the signed area between the graph of the constant function and the x-axis over the interval of integration.

Proof

I can derive the formula for the indefinite integral of a constant using the linearity of integration, specifically the constant multiple rule.

This rule states that a constant factor can be moved outside an integral.

$$ \int k \cdot f(x)\,dx = k \int f(x)\,dx $$

To apply this rule, I first rewrite the constant \(k\) as the product \(k\cdot 1\).

$$ \int k\,dx = \int k \cdot 1\,dx $$

Now I can take \(k\) outside the integral.

$$ \int k\,dx = k \int 1\,dx $$

The remaining integral is straightforward. Since the derivative of \(x\) is 1, an antiderivative of the constant function \(f(x)=1\) is \(F(x)=x\).

Therefore,

$$ \int k\,dx = kx+C $$

This proves that the indefinite integral of a constant is the constant multiplied by the variable of integration, plus an arbitrary constant of integration.

In other words, the linearity of integration allows me to derive the general formula directly from the integral of \(f(x)=1\).

Proof of the Formula for the Definite Integral of a Constant

Now consider the constant function \(f(x)=k\) over the interval \([a,b]\), where \(a< b\). Since every constant function is continuous, I can apply the standard properties of definite integrals.

I want to evaluate

$$ \int_a^b k\,dx $$

First, I use the linearity of the definite integral, which allows a constant factor to be taken outside the integral.

$$ \int_a^b k\cdot f(x)\,dx=k\int_a^b f(x)\,dx $$

Writing the constant function as \(k\cdot 1\), I obtain

$$ \int_a^b k\,dx=k\int_a^b 1\,dx $$

So the problem comes down to evaluating the definite integral of the constant function \(f(x)=1\).

To do this, I apply the Fundamental Theorem of Calculus, specifically its evaluation formula, also known as the Newton-Leibniz formula.

The theorem states that if \(f(x)\) is continuous on \([a,b]\) and \(F(x)\) is an antiderivative of \(f(x)\), then

$$ \int_a^b f(x)\,dx=F(b)-F(a) $$

Here, an antiderivative \(F(x)\) is a function whose derivative is \(f(x)\).

For \(f(x)=1\), I can choose \(F(x)=x\), because the derivative of \(x\) with respect to \(x\) is 1.

Applying the theorem gives

$$ \int_a^b 1\,dx=[x]_a^b=b-a $$

Finally, I substitute this result into the expression obtained using the constant multiple rule.

$$ \int_a^b k\,dx=k(b-a) $$

This completes the proof. The definite integral of a constant equals the constant multiplied by the length of the interval of integration.

Alternative Proof Using the Mean Value Theorem for Integrals

There is another way to derive the same formula, using the Mean Value Theorem for Integrals.

This theorem states that if a function \(f(x)\) is continuous on a closed interval \([a,b]\), with \(a< b\), there is at least one point \(c\) in the interval such that

$$ \int_a^b f(x)\,dx=f(c)(b-a) $$

In simple terms, the theorem tells me that the definite integral of a continuous function equals the function's value at some point in the interval multiplied by the interval's length.

For the constant function \(f(x)=k\), the situation is particularly simple. The function has the same value at every point, so

$$ f(c)=k $$

Substituting this into the Mean Value Theorem for Integrals, I immediately get

$$ \int_a^b k\,dx=k(b-a) $$

This is exactly the same formula obtained using the linearity of integration and the Fundamental Theorem of Calculus.

There is also an interesting detail. The Mean Value Theorem for Integrals guarantees the existence of at least one point \(c\) for which the formula holds. For a constant function, however, the equality holds at every point in the interval because the function's value never changes.

And so on.