Question

Will somebody explain integrals to me?

Answer 1

take calc II

Answer 2

Hey help me, don't rub it in. Explain it to me, if you don't have something even slightly insightful to say don't say it.

Answer 3

integral gives you area under a curve.

The simplest way to start is to draw a graph of a horizontal line, like y = 2.

Integrating that line "curve" gives you the area under it. Integrating it from say, x=0 to x=1, gives you a rectangle that is 1 unit long by 2 units high, so area is 2.

However, integrals allow you to find area under any curve, not just straight lines.

Answer 4

I am not old enough to take calc, I am only 14, but I can do trig.

Answer 5

Thank you

Answer 6

trig isn't quite enough for integrals...

Answer 7

Cool, now you understand integrals. it's just that easy.
Why would you think some random people commenting on some random forum would be a useful substitute for a challenging 12 week course?

Answer 8

@Plutog55
do you want to know how to "do integrals" or just understand what they are?

Answer 9

@Algebraic! is right... you are asking for an explanation that is easily a whole class on its own.

Answer 10

But "area under a curve" is a good quick idea of what an integral means.

The opposite of an integral is a derivative.. .also beyond trig... a derivative gives the slope of a curve at any point along the curve. Integrating gives the area under the curve, either in general or over some range of x values.

Answer 11

did I say I won't take the class now? no, I am not stupid. I know trig isn't enough for that. It's just some curiosity, It is in no way a replacement. No harm done, and yes @JakeV8 I would love to know how.

Answer 12

I can't teach you the everything about integrals, but here's a shortcut technique you can play around with.

Take a function like f(x) = 2. This is the y=2x line I mentioned earlier.

To integrate, you need to think of a new function that has an exponent on the x-term that is 1 larger than the current function. In this function, it is really equivalent to saying f(x) = 2 = 2x^0

So the integral of f(x) will have an exponent of 1... x^1.

The other part is that the exponent on the integrated term (in this case 1), ends up as a coefficient in the original term, prior to integrating. For this case, since multiplying by 1 doesn't change anything, you don't see the effect, but for x^2, x^3, and higher, it does make a difference.

This is confusing, I'm sure, so let's look at a couple of examples to see the pattern.

Answer 13

1st, integrating f(x) = constant
f(x) integral of f(x)
2 2x
3 3x

2nd, integrating f(x) = mx i.e. ,a line
f(x) = integral of f(x)
x (1/2)x^2 <<-- exponent goes from 1 to 2
^ <<-- 2 exponent on x^2 multiplies by coefficient (1/2)
^<<<<<<<< to leave x with a coefficient of 1 on the "un-integrated" function

Answer 14

hmmm.....

Answer 15

f(x) integral of f(x)
2x x^2 <<-- same idea as for f(x)=x... now the 2 coefficient doesn't have the (1/2) to cancel, so it shows up in the original f(x) = 2x as the coefficient "2"

Answer 16

f(x) = x^3

integral of f(x) = (1/4)x^4

since to get back to f(x), you would first multiply the exponent 4 by the (1/4) and then reduce the exponent to 3, leaving x^3.

Answer 17

The reason this isn't very obvious yet is that typically you learn the opposite of integrals first... you learn differentiation or "taking derivatives".

the derivative of x^2 is found by multiplying by the exponent "2", and then reducing the exponent by 1...

derivative of x^2 = 2x

Integrals are the opposite...

integral of 2x = x^2
then to get back, take the derivative of x^2 to get 2x back again

Answer 18

Ah! OK... I think I get it now. Thank you very much.