Lean Calculus

Our study of integrals was motivated by the area problem so it feels necessary to at least note that in a fairly broad sense, we’ve solved the area problem. Consider, for example, $f(x)=\sqrt{x}$ and $g(x)=x^3\text{.}$ Let’s graph them both together. Take a look at that oddly shaped area trapped between the two curves. Can we calculate that area? Sure. We have the skills now. Notice that the area occurs over the interval $0\le x\le 1\text{.}$ The curve on top is $f(x)=\sqrt{x}\text{,}$ An antiderivative of $\sqrt{x}$ is $2x^{3/2}/3$ so that according to the FTC, the area under that curve is ${\int }_0^1\sqrt{x}dx=2(1{)}^{3/2}/3-2(0{)}^{3/2}/3=2/3\text{.}$ The curve on the bottom is $g(x)=x^3\text{,}$ An antiderivative of $x^3$ is $x^4/4$ so that according to the FTC, the area under that curve is ${\int }_0^1{x}^{3}dx=(1{)}^4/4-(0{)}^4/4=1/4\text{.}$ So, the area trapped between is $2/3-1/4=8/12-3/12=5/12\text{.}$ -- very quick to calculutate with the skills we now have.