Evaluate the indefinite integral:
∫9dx= + C.

Question

Evaluate the indefinite integral: 
∫9dx=   + C.

zepdrix · Answer

The indefinite integral of a constant? Hmm..

What's the derivative of $\large x$?  Just $\large 1$ right?
How bout the derivative of $\large 9x$? Hmm $\large 9$ yes?

So integrating $\large 9$ should give us??

zepdrix · Answer

If you're not quite convinced, you could also try applying the `Power Rule for Integration`. You could rewrite your integral like this,$$\large \int\limits 9x^0dx$$Then applying the power rule gives us,$$\large \frac{9x^{0+1}}{0+1}+C \qquad = \qquad 9x^1+C$$

anonymous · Answer

ok so i wasnt in class so how do we get this answer

anonymous · Answer

@zepdrix

zepdrix · Answer

Integration is the opposite process of taking a derivative.
Another word that you might here is "Anti-differentiation".

$$\large \int\limits 9\;dx$$
To evaluate this integral, we'll have to find the anti-derivative.
When we integrate, the big swirly bar and the dx will disappear as a part of this process.
$$\large \int\limits\limits 9\;dx \quad = \quad 9x$$See how the dx and swirly S disappeared?
The derivative of $\large 9x$ is $\large 9$, therefore the anti-derivative of $\large 9$ must be $\large 9x$.
Make a little bit of sense? D:

anonymous · Answer

ah ok ya that makes sense

zepdrix · Answer

There is a +C is for another reason.

Here is a quick example:
$$\large f(x)=x^2+7$$$$\large f'(x)=2x$$
Let's look at another similar function.$$\large g(x)=x^2+33$$$$\large g'(x)=2x$$

So if I were to ask you to evaluate this integral,$$\large \int\limits 2x\;dx$$I'm asking you to find the anti-derivative of this function ~ The function that if we took the derivative of, would give us 2x.
We have a problem though, is the anti-derivative going to be, $\large x^2+7$, or $\large x^2+33$ ???

The anti-derivative represents a FAMILY of solutions. It's the entire collection of possible x^2's.

So we throw the +C on the end to show that the anti-derivative could be $\large x^2+7$ or $\large x^2+392$ or any unknown constant that would leave us with 2x if we took the function's derivative.

anonymous · Answer

ok that makes alot more sense

anonymous · Answer

thank you so much

zepdrix · Answer

c: