Find the Antiderivative of the function

[femmed0ll]

Find an antiderivative of the given function?
1) x^(-4) + 1 / 3?x
2) cos pi x + 7 sin x/7
what are the steps to solve these equations? im soo lost..

 

[galactus]

Find an antiderivative of the given function?

Hi there. These are not equations, they are more like expressions. An equation has an equal sign and an expression does not.

Anyway, just clearing that up a little for you.

1) \(\displaystyle \int(x^{-4} + \frac{1}{3\sqrt{x}})dx\)

This is the opposite of differentiation, so with these problems, do the opposite of what you would do when you find the derivative. That is a simplistic way to explain it. Add 1 to the exponent and divide into the coefficient.

Rewrite as:

\(\displaystyle \int x^{-4}dx+\frac{1}{3}\int x^{\frac{-1}{2}}dx\)

This is very straightforward with no fancy tricks. Can you take it from here?.

2) \(\displaystyle cos(\pi x) + 7sin(\frac{x}{7})\)

\(\displaystyle \int cos(\pi x)dx+7\int sin(\frac{x}{7})dx\)

One thing to learn is that the antiderivative of cos(x) is sin(x). For the first part, since we have the pi*x inside, we can make a simple substitution.

Let \(\displaystyle u={\pi}x, \;\ du={\pi}dx, \;\ \frac{du}{\pi}=dx\)

See?. Now make the subs and we get \(\displaystyle \frac{1}{\pi}\int cos(u)du=\frac{sin(u)}{\pi}\)

Now, resub back in. Note that since \(\displaystyle u={\pi}{x}\), we have \(\displaystyle \frac{1}{\pi}sin({\pi}x)\). That's it.

Now, can you do the other part of the problem and let me know what you get?.

 

[femmed0ll]

thanks so much...i got -49 cos x/7 !
if thats right..

 

[galactus]

Yep. That's it. Remember, that is oart of that second problem.

Altogether, it is \(\displaystyle \frac{sin(\pi x)}{\pi}-49cos(\frac{x}{7})+C\)

The C is an arbitrary constant. Some instructors may not care if you attach that, but better safe than sorry.

Good work. You'll get the hang of it for your test tomorrow. Practice problems in the book.