∫√x(-x^7-x^4)

Question

∫√x(-x^7-x^4)

zepdrix · Answer

Is everything under the root? Or just the first x?

anonymous · Answer

First x

zepdrix · Answer

First off, $\large \bf\color{royalblue}{\text{Welcome to Open Study! :)}}$


Let's start by rewriting our root as a rational expression.
$$\large \int\limits \sqrt x \left(-x^7-x^4\right)dx \qquad=\qquad \int\limits x^{1/2}\left(-x^7-x^4\right)dx$$
From here we can distribute our x^1/2 to each term in the brackets.
To multiply each x, we'll use rules of exponents.

zepdrix · Answer

Is the fancy math stuff showing up ok for you?
If you're using Internet Explorer you might just see a bunch of jumbly code.

zepdrix · Answer

$$\large \int\limits -x^{(7+1/2)}-x^{(4+1/2)}dx$$

anonymous · Answer

Its fine

zepdrix · Answer

When we multiply terms of similar bases, we `add` the exponents.
Which gives us something like...$$\large \int\limits -x^{15/2}-x^{9/2}\;dx$$Confused by any of that? :o

anonymous · Answer

For the second step you multiplied the X^1/2 in right?

zepdrix · Answer

Ya :)
$$\large \int\limits\limits x^{1/2}\left(-x^7-x^4\right)dx \qquad=\qquad \int\limits -x^{7}x^{1/2}-x^4x^{1/2}\;dx$$

anonymous · Answer

Ok

zepdrix · Answer

From here,$$\large \int\limits\limits -x^{15/2}-x^{9/2}\;dx$$we would just apply the Power Rule for Integration.
Lemme know if those fraction give ya any trouble.

anonymous · Answer

No, I'm fine. Thanks for the help