Indefinite Integrals - Problem 2

Explanation

One useful property of indefinite integrals is the constant multiple rule. This rule means that you can pull constants out of the integral, which can simplify the problem. For example, the integral of 2x + 4 is the same as the 2 multiplied by the integral of x + 2. However, it is important that only constants—not variables—are pulled out of the integral.

Another property is the power rule of antiderivatives (remember that to take the integral of something is to antidifferentiate). The power rule states that the antiderivative of a xn is xn+1/n + c, where c is some constant.

Remember that roots are fractional powers. For instance, the square root of x is x1/2.

Transcript

Let's solve some harder indefinite integrals. Now we're going to need to use the all the tools we have here, both the properties; the constant multiple rule, and the sum rule. Also this formula, this is the power rule for antidifferentiation. How to antidifferentiate x to the n. So we'll need all three of these.

Let's start by recognizing that the fourth root of x, which I have to integrate first, is actually a power of x, so you should write it as such. Maybe that's the first thing you should do, is write 5x, and that's the ¼ power. Then we're going to pull the 5 out. That's an application of the constant multiple rule. If you have a constant in front of your function, you can pull that constant out of the integral. So 5 times the integral of x to the ¼dx.

Now I have an x to the n. I can just use this power rule. This becomes 5 times 1 over ¼ plus 1 times x to the x to the ¼ plus 1, plus, and I'll call it c1. Now ¼ plus 1 is 5/4. 1 over 5/4 is 4/5. So this is 5 times 4/5x to the 5/4 plus 5 times c1. The 5 times 4/5, 5's cancel, I get 4x to the 5/4, plus, and I'm going to rename this c. 5c1 will become just c. So that's my answer 4x to the 5/4 plus c.

Now I'm going to check this using differentiation. The problem asks me to, but even if your teacher doesn't ask you to check by differentiating, you might impress them by doing this. So it's a good idea to check. It's also a good habit to get into just in case you're making mistakes.

So, check by differentiating. 4x to 5/4 plus c. So the derivative is 4 times the (5/4 comes down) x to the (I subtract 1 from the 5/4) I get ¼. Then the derivative of the plus c, that's just 0. 4 times 5/4 is 5x to the ¼. That's it. That's my original function. So it checks. This answer is correct. These are my antiderivatives of 5x to the ¼.

Let's take a look at a slightly harder derivative. At the integral of root x times x plus 1 over x. Now it's really tempting to invent properties of integrals that don't exist. For example, you may want to antidifferentiate this function; root x, and these two separately and write the product. That doesn't work with integrals. There is no product rule for integrals. So instead, what you should do is combine these two. These are all just powers of x. By multiplication, you can combine them. So I would convert everything to powers of x first. X to the 1/2, x to the 1 plus x to the -1. Then you combine these. Just distribute the x to the 1/2 through these two terms.

X to the 1/2 times x, is going to be x to the 3/2. X to the 1/2 times x to the -1 is x to the -1/2dx. I have to separate this. Let's not forget we have to use the sum rule here. I'm separating this integral of a sum into a sum of integrals. So I can treat each of these integrals separately. I have x to the 3/2 here, x to the -1/2 here.

Now I have something that looks exactly like my power rule. I'm integrating something of the form x to the n. So the rule says add 1 to the exponent. 3/2 plus 1 is 5/2. So we have x to the 5/2 divided by 5/2. Now strictly speaking, when I integrate this guy, I get a constant. When I integrate this guy, I'll get another constant and I'll use the same rule. I add 1 to the exponent. When you add 1 to -1/2 you get +1/2. So it's x to the 1/2 over 1/2 plus another constant, c2.

Now remember, dividing by a fraction is like multiplying by its reciprocal. So dividing by 5/2 is like multiplying by 2/5. So 2/5 times x to the 5/2 plus, I'm going to combine these cs at the end. Dividing by 1/2 is like multiplying by 2. So this is 2x to the 1/2. Then the c1 plus c2, I'm going to combine into a single constant c. That's my answer. These are my antiderivatives of this function.

So let's check that by differentiating. So let's take it up here. The derivative of, I have 2/5 x to the 5/2 plus 2x to the 1/2 plus c. So first, this term; 2/5 times, and the derivative of x to the 5/2, is 5/2x to the (and I subtract 1 from the exponent) 5/2 minus 1 is 3/2. Plus 2 times (and I differentiate x to the 1/2) the 1/2 comes down in front. I have x to the 1/2 minus 1, -1/2. Of course the derivative of the plus c is 0. So 2/5 times 5/2 is 1. I get x to the 3/2, 2 times 1/2 a is 1. So I get x to the -1/2.

Let's compare this to what we had in the beginning. Easiest to compare to this guy here. That looks correct. So my derivative checks. These are the antiderivatives of root x times the quantity x plus 1 over x.

Tags
indefinite integral antiderivative antidifferentiation differentiation