That first question probably had a misprint. The new one can be done.

Make the substitution u = x-2

Differentiate: du = dx. (Strictly, this is poor terminology but it works )

When x = -1, u = -3 and when x = 4, u = 2

So

]]>This is the new one:

Is this statement valid?I've worked a little on it before getting stuck at:

]]>It is possible to find some answers but I am unable to find a general solution.

]]>I don't understand how this question can be done. Let me explain. Perhaps someone wiser than me will show why I'm wrong. Here goes:

You are not told very much about the function h. All we know is a certain integral. It's the coloured area on this graph.

Now I've made up a shape for h. We don't know the true shape, but my example will do for explanation purposes.

Then you are asked about the function 3h(x-2). Now the 3 is no problem. You get the integral of h(x-2) and multiply it by 3. So I'll assume we are ok to do that at the end and concentrate on the integral of h(x-2)

So what is this graph like? h(x-2) when x = 2 will be h(2-2) = h(0). So when its x coordinate is 2 the y coordinate will be the original graph's y coordinate at x = 0

Similarly at x = 6, h(x-2) = h(6-2) = h(4) so the y coordinate at x=6 will be the same as the y at x=4 on the original graph. Every point just moves across 2 units to the right. I've shown that graph too. Now we can say nothing about that graph outside the interval [2,6]

But you are asked to say what the integral will be from 1 to 5. How can you ? when you don't know the shape of the graph to the left of x = 2 and you don't know how much area is omitted between 5 and 6.

I can show you either of these:

and

Beyond that I cannot say.

Bob

]]>Do i make it du=hdx]]>

Where does it say that h(u) = h(x) = 5?

But first what did you calculate the new limits of integration as?

]]>and then add in the limits?

]]>Where would I go from there?]]>