# 7c. Find *y* given *dy/dx*

## Integration Mini Video Lecture

This is a common integral calculus question, where we are given an expression for *dy/dx* and *x- *and *y*-values, and we need to find ans expression for *y*.

## Video transcript

[Music...]

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A very common integration problem is [when] we are given an expression for `dy/dx` and we are given a point that the curve passes through and we need the find the original function.

This is integration because we are told the `dy/dx` and we need to find the original expression.

Ok, ...so if I've got `dy/dx` equals 3*x*^{2} + 2. I recognise that what I need to do is to integrate this thing first.

So I write *y* equals the integral of bracket 3*x*^{2} + 2 *dx* .

Now how did I know that, becuase `dy/dx` is 3*x*^{2} + 2, *y* must be the integral of 3*x*^{2} + 2.

`dy/dx` [differentiation] and integration are opposite processes.

Ok, so I just integrate this and I get `x^3/3`, and [there is] 3 on the front, so the 3 cancels out so I just get *x*^{3} + 2x + *K*.

This "plus *K*" is very important. Please don't forget it, because it's very important for this sort of problen. In fact very important for all indefinite integral problems.

Ok great, now what do I do?

I am told that the curve passes through (1, 4), and if the curve passes through (1, 4), it means when *x* = 1, then *y* must equal 4.

So, who cares?

We can use this thing we found here to find the value of *K*, and then we can expresss the exact answer for the function *y* equals function *x*

OK, subsitituting this into this, on the left hand side we are going to have 4, and on the right hand side, we are going to have (1)^{3} and then plus 2 times 1 plus *K*.

Now what am I doing, I am claiming up here that *y = x*^{3} + 2x + *K. *I know that when *x* is 1, *y* is 4, so I substitute *y* equals 4 into here. There it is. And I substitute *x* equals 1 into here and here.

And so now this just simply becomes...

4 = 1 + 2 + *K*.

Well that's pretty easy...

*K* equals 1... [mistake, sorry!]

OK, so now I can re-express this thing I have on the top here, and go like this...

Just copy this bit and paste... plus 1.

I found *K* equals 1 and...

I will just put *K* equal 1 here and *y* equals *x*^{3} + 2x + *K* and put it down here.

So I am finished. This is the answer.

This is the function *y = x*^{3} + 2x + 1.

Its derivative is 3*x*^{2} + 2.

And it passes through the point (1,4).

And now for the graph of the function,

We can see that it passes through (1,4).

We were given the `dy/dx` expression and we have found the "*y = * " expression and we can graph it. And it's a cubic.

[Music...]

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