In this lesson, we'll try to find the area underneath any function \(kx^m\) where \(k\) and \(n\) are any constants such that \(n≠-1\). In other words, we want to solve the integral

$$\int{kx^m}dx.$$

In the lesson, Overview of Single-Variable Calculus, we got the sense that integrals and derivatives are, somehow, opposites of one another. Let me give an example of that. Suppose that we wanted to find the derivative \(f'(x)\) or \(f(x)=x^2\). To do that, we could just use the following definition of the derivative that we derived in a previous lesson:

$$f'(x)=\lim_{Δx→0}\frac{f(x+Δx}{Δx}.$$

Since we already solved this problem in a previous lesson, we won't evaluate this limit here. According to our calculation in a previous lesson on derivatives, \(f'(x)=2x\). But now, suppose that I asked you to find \(\int{f'(x)dx}=\int{2x}dx\) - in other words, find the area underneath the function \(2x\). As we discussed in the lesson, *Overview of Single-Variable Calculus*, using integrals (or derivatives) are only necessary when the slope of the curve keeps changing. But as we can see in Figure 1, the slope of the function \(2x\) is just the constant \(2\). Finding the area underneath this function is simply a matter of finding the area of a triangle. The width of the orange triangle in Figure 1 is \(x\) and its height is \(2x\); thus the area of the triangle must be \(A=\frac{1}{2}(x)(2x)=x^2\). Since the area of the triangle is equal to the area underneath \(f'(x)=2x\), and since the area underneath \(2x\) is just \(\int{2xdx}\), it follows that

$$\int{f'(x)dx}=\int{2xdx}=x^2.\tag{1}$$

The point that we need to focus on here is this: the derivative took us from \(f(x)\) to \(f'(x)\), and the integral took us from \(f'(x)\) back to \(f(x)\) again. The integral "undid the derivative," so to speak. This is why the integral is also commonly refereed to as the anti-derivative.

Using the definition of a derivative, we showed in a previous lesson that the derivative \(\frac{d}{dx}(kx^m)\) is just

$$\frac{d}{dx}(f(x))=\frac{d}{dx}(kx^m)=kmx^{m-1}=f'(x).$$

In that lesson, we used the derivative to go from \(f(x)\) to \(f'(x)\). But the integral \(\int{f'(x)dx}=\int{kmx^{m-1}dx}\) undoes that derivative and gives us \(kx^m\) again. Thus,

$$\int{kmx^{m-1}dx}=kx^m.\tag{2}$$

The inside of the integral, \(kmx^{m-1}\) (also called the integrand), looks a little sloppy. But the term \(km\) is just some constant which we'll represent by \(a\). Also, let's define the constant \(n\) as \(n≡m-1\). Thus, \(m=n-1\). Substituting \(km=a\), \(n=m-1\), and \(m=n-1\) into Equation (2) we have

$$\int{ax^ndx}=\frac{a}{n+1}x^{n+1}.\tag{3}$$

For any function \(f(x)=ax^n\) where \(a\) can be any constant and where \(c\) can be any constant such that \(n≠-1\), to find the integral o fthat function all we have to do is the following procedure: add \(1\) to the exponent of \(x\), then divide by the new exponent.

**Example 1**

Suppose that I asked you to solve the integral

$$\int{3x^2dx};$$

in other words, I want you to find the area underneath the function \(f(x)=3x^2\). In this problem, we can just pattern match to see that \(a=3\) and \(n=2\). According to Equation (3), the answer to this problem must be

$$\frac{a}{n+1}x^{n+1}.$$

Substituting \(n=2\) and \(a=3\), we find that

$$\int{3x^2dx}=\frac{3}{2+1}x^{2+1}=\frac{3}{3}x^3=x^3.$$

Thus, the area underneath \(3x^2\) is just \(x^3\). If we took the derivative of \(x^3\), we would have

$$\frac{d}{dx}x^3=3x^{3-1}=3x^2.$$

Thus, by taking the derivative, we'd get the original function back as expected.

**Example 2**

$$\text{Solve the integral }\int{105x^{π+5}dx}.$$

The integral above might look pretty sloppy, but we'll see that hopefully this problem isn't too bad. In this problem, \(a=105\) and \(n=π+5\). The solution to this integral is just

$$\int{105x^{π+5}dx}=\frac{105}{(π+5)+1}x^{(π+5)+1}=\frac{105}{π+6}x^{π+6}.$$

Thus, the solution to the integral is just

$$\int{105x^{π+5}dx}=\frac{105}{π+6}x^{π+6}.$$