I have written "… " where I would imagine the student is replying. I have then provided prompts assuming that the student did not come up with the answer straight away. Integration can be useful in lots of areas like mechanics (finding velocities/accelerations), probability or simply for calculating areas of strange shapes. Would you like to see an example? Imagine you're a farmer and you want to know how much food your trough can hold. This is what it would look like (draw a trough). 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 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 to 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.