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# 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

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### Related lessons on IntMath

Antiderivatives and The Indefinite Integral

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 3x2 + 2. I recognise that what I need to do is to integrate this thing first.

So I write y equals the integral of bracket 3x2 + 2 dx .

Now how did I know that, becuase dy/dx is 3x2 + 2, y must be the integral of 3x2 + 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 x3 + 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 = x3 + 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 x3 + 2x + K and put it down here.

So I am finished. This is the answer.

This is the function y = x3 + 2x + 1.

Its derivative is 3x2 + 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.

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