Question

integrate (x^3+x)/x^2 from 1 to 3

Answer 1

\[\large \int\limits_1^3 \frac{x^3+x}{x^2}dx\]We can split this into two fractions,\[\large \int\limits\limits_1^3 \frac{x^3}{x^2}+\frac{x}{x^2}dx\]Then simply divide each term to simplify, and the apply the `Power Rule for Integration`.
Understand how to proceed? :)

Answer 2

no, unfortunately i don't know how to integrate. do i sub in 1, 2 and 3 for each x?

Answer 3

After you've simplified you should have,\[\large \int\limits\limits\limits_1^3 x+\frac{1}{x}dx\]

We'll integrate these separately, the same way we would do if we had to take a derivative with addition between the terms. The steps will be a little different though.

Answer 4

Remember the `Power Rule for Derivatives`?

It involves 2 steps:
~Multiply by the power.
~Decrease the power by one.

The `Power Rule for Integration` will be the opposite, and in the opposite order.
~Raise the power by 1.
~Divide by the power.

Answer 5

Here's a quick example,\[\large \int\limits x^3\;dx \qquad =\qquad \frac{x^{(3+1)}}{3+1} \qquad = \qquad \frac{1}{4}x^4\]

Answer 6

So in your problem, that first term is \(x^1\).
Try applying this rule to it.

Answer 7

so would it look like this? \[\int\limits_{1}^{3}x+0 dx\]

Answer 8

We can't apply the `Power Rule for Integration` to that second term. It's a special power that we're not allowed to apply the rule to.

Only apply it to the first term.
Did you try? :o

Answer 9

yes. does that mean that it would be \[1/2\int\limits_{1}^{3}x^2\]

Answer 10

Yes good c:

Now for the second term, we'll need to think back to an important derivative.\[\large \left(\ln x\right)'=\frac{1}{x}\]The `derivative` of ln x gives us this.
So if we want the `Anti-derivative` of 1/x, what will we get? :)

Answer 11

is it lnx? lol

Answer 12

yessss c:

Answer 13

great! ok now what do i do with the 3 and 1? do i sub in now?

Answer 14

Remember when you would take derivatives, the derivative operator, or the prime, would disappear as a part of the process of finding the derivative.

Same thing happens with integration, the Swirly S bar, and the dx will disappear as a part of the integration process.

We'll show that we still need to plug in our limits by drawing a bar behind our anti-derivative that we've found.\[\large \int\limits\limits\limits\limits_1^3 x+\frac{1}{x}dx \qquad = \qquad \frac{1}{2}x^2+\ln x \;|_1^3\]

Answer 15

So we'll plug in the 3, and subtract from that, the function with 1 plugged in.\[\large \frac{1}{2}x^2+\ln x \;|_1^3 =\]

\[\huge \left(\frac{1}{2}\color{royalblue}{3}^2+\ln \color{royalblue}{3}\right)-\left(\frac{1}{2}\color{royalblue}{1}^2+\ln \color{royalblue}{1}\right)\]

Answer 16

so is the answer\[4+\ln 2\]

Answer 17

\[\large \ln1=0\]

Answer 18

oh right! so the answer is just 4?

Answer 19

Where did your ln3 go? :D
\[\large 4+\ln3\]

Answer 20

oh ok. i understand more now. thanks so much!

Answer 21

yay c: