Page:Calculus Made Easy.pdf/229

 will be read: find the integral of $$y \cdot dx$$ between the inferior limit $$x_1$$ and the superior limit $$x_2$$.

Sometimes the thing is written more simply

Well, but how do you find an integral between limits, when you have got these instructions?

Look again at Fig. 52 (p. 206). Suppose we could find the area under the larger piece of curve from $$A$$ to $$Q$$, that is from $$x=0$$ to $$x=x_2$$, naming the area $$AQNO$$. Then, suppose we could find the area under the smaller piece from $$A$$ to $$P$$, that is from $$x=0$$ to $$x=x_1$$, namely the area $$APMO$$. If then we were to subtract the smaller area from the larger, we should have left as a remainder the area $$PQNM$$, which is what we want. Here we have the clue as to what to do; the definite integral between the two limits is the difference between the integral worked out for the superior limit and the integral worked out for the lower limit.

Let us then go ahead. First, find the general integral thus:

and, as $$y=b+ax^2$$ is the equation to the curve (Fig. 52),

is the general integral which we must find.

Doing the integration in question by the rule (p. 196), we get