I have written "..." where I imagine the student is replying. I have then provided prompts assuming that the student did not come up with the answer straight away. The scenario could be very different if the student was very competent in the area.
Integration can be useful in lots of areas like mechanics (finding velocities/accelerations), probability or simply for calculating areas of strange shapes. Let's take an example. Imagine you're a farmer and you wanted to know how much food your trough could hold. As it's a prism, you could calculate its volume by multiplying the area of its cross-section by its length. What shape could you use to describe the cross-section? A parabola, yes! Now say we wanted to calculate the area of this cross-section. How might we go about it?... Why don't you try drawing the shape on a graph, and shading the region that describes the trough's cross-section? … Does that look like the area under a parabola (eg: -x^2 +4)? Ok, do you remember calculating the area under a parabola in class? Let's take a look at your notes.... Here it says that the integral of x^n=x^{n+1}/(n+1)+C. Could you remind me how this was obtained? (Either the student gives an understanding of the relationship between differentiation and integration, or I will describe it if I feel it's necessary). Great, so how could you apply this to our problem? I see you've noted that n=2 for the parabola and that the limits of integration are -2 and 2. That's good! Could you give me an exact answer? Yes, 32/3 is correct!
Example answer:First solve -x^2+4=0 to get x=2,-2. This tells us to integrate -x^2+4 from -2 to 2. So we obtain -2^3/3+42-(2^3/3-42)=-16/3+16=32/3.