Integrals in Calculus?

<p>...I just don't get them. Anyone care to explain?
I always do half of a problem and think it's the end, but then I find out there's more. So annoying. How do I know when I'm really done?</p>

<p>Integral = antiderivative.</p>

<p>What is the derivative of x^2?
2x</p>

<p>What is the integral of 2x
x^2 + C</p>

<p>You’re done once you’re back to the original form of the problem plus C (the constant of integration).</p>

<p>does the constant differ for each problem? (clearly I haven’t been paying attention in class)</p>

<p>oh and when you integrate from A–>B, that’s where I get confused as to what I should do</p>

<p>Integrating from A —> B just means finding the area under said function from A —> B. Just use the fundamental theorem of calc to find the integral, find the antiderivative F of the function f(x) in the integral, then just do F(B)-F(A).</p>

<p>after you antidifferentiate your function, to find the definite integral from A to B, you have to do f(B) - f(A). So for instance you want to find the definite integral of 3x^2 from 1 to 2. You first antidifferentiate the function which gives you x^3. Then you do f(2) - f(1) which is equal to 7. it can get way more complicated with constants out in front and other stuff, but this is the most standard form.</p>

<p>ahh okay. makes sense now</p>

<p>Referring to the constant if you are not given limits (i.e. not integrating between A and B) you have to add the +c (the constant of integration). This is infinately many functions differentiate to the same thing for example x^2 and x^2 + 1 etc. all differentiate to 2x. The constant of integration compensates for this.</p>

<p>Note: This idea works in multivariable/vector calc as well but is a function of other other variables rather than a constant.</p>