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Area Problem

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Suppose that we are given continuous function y=f(x)y={f{{\left({x}\right)}}} on [a,b]{\left[{a},{b}\right]} such that f(x)≥0{f{{\left({x}\right)}}}\ge{0} for all area problemx∈[a,b]{x}\in{\left[{a},{b}\right]}.

We want to find area S{S} that lies under curve f(x){f{{\left({x}\right)}}} and bounded by lines x=a{x}={a}, x=b{x}={b} and x-axis.

Before doing this imagine right bound b{b} is not fixed, we can move it. Then for different values of right bound we will obtain different values of area, so we can consider area as function P(x){P}{\left({x}\right)} of right bound. In particular, P(a)=0{P}{\left({a}\right)}={0} (because area from a{a} to a{a} is 0) and P(b)=S{P}{\left({b}\right)}={S}.

To get a better understanding what is P(x){P}{\left({x}\right)} consider next example.

Example 1. Consider function y=x−1{y}={x}-{1}. If P(x){P}{\left({x}\right)} represents area of figure under y=x{y}={x}, above x-axis on interval [1,x]{\left[{1},{x}\right]} find P(1){P}{\left({1}\right)}, P(2){P}{\left({2}\right)}, P(3){P}{\left({3}\right)} and P(4){P}{\left({4}\right)}.varying area

P(1){P}{\left({1}\right)} is area on interval [1,1]{\left[{1},{1}\right]}, so it is 0 (because length of interval is 0).

P(2){P}{\left({2}\right)} is area on interval [1,2]{\left[{1},{2}\right]}. It is area of right-angled triangle with legs 1 and 1. So, P(2)=12⋅1⋅1=12{P}{\left({2}\right)}=\frac{{1}}{{2}}\cdot{1}\cdot{1}=\frac{{1}}{{2}}.

P(3){P}{\left({3}\right)} is area on interval [1,3]{\left[{1},{3}\right]}. It is area of right-angled triangle with legs 2 and 2. So, P(3)=12⋅2⋅2=2{P}{\left({3}\right)}=\frac{{1}}{{2}}\cdot{2}\cdot{2}={2}.

P(4){P}{\left({4}\right)} is area on interval [1,4]{\left[{1},{4}\right]}. It is area of right-angled triangle with legs 3 and 3. So, P(4)=12⋅3⋅3=92{P}{\left({4}\right)}=\frac{{1}}{{2}}\cdot{3}\cdot{3}=\frac{{9}}{{2}}.

As can be seen P(x){P}{\left({x}\right)} is indeed function: for any particular value of x{x} there is corresponding P(x){P}{\left({x}\right)}.

varying area

Now, suppose we want to find derivative of P(x){P}{\left({x}\right)}. For this increase x{x} by Δx\Delta{x}. Corresponding increase in area is ΔP=P(x+Δx)−P(x)\Delta{P}={P}{\left({x}+\Delta{x}\right)}-{P}{\left({x}\right)} (figure ACEF).

If we denote maximum and minimum value of f(x){f{{\left({x}\right)}}} on interval [x,x+Δx]{\left[{x},{x}+\Delta{x}\right]} by M{M} and m{m} respectively, then area of figure ACEF is less than area of rectangle ACDF and greater than area of rectangle ABEF.

Since area of rectangle ACDF is MΔx{M}\Delta{x} (height multiplied by base)and area of rectangle ABEF is mΔx{m}\Delta{x} then

mΔx<ΔP<MΔx{m}\Delta{x}<\Delta{P}<{M}\Delta{x} or m<ΔPΔx<M{m}<\frac{{\Delta{P}}}{{\Delta{x}}}<{M}.

Now let Δx→0\Delta{x}\to{0}. Since f{f{}} is continuous then both m{m} and M{M} approach f(x){f{{\left({x}\right)}}}.

Thus, by Squeeze Theorem lim⁡Δx→0ΔPΔx=f(x)\lim_{{\Delta{x}\to{0}}}\frac{{\Delta{P}}}{{\Delta{x}}}={f{{\left({x}\right)}}}.

But we recognize in lim⁡Δx→0ΔPΔx\lim_{{\Delta{x}\to{0}}}\frac{{\Delta{P}}}{{\Delta{x}}} derivative of P(x){P}{\left({x}\right)}, so P′(x)=f(x){P}'{\left({x}\right)}={f{{\left({x}\right)}}}.

In other words P(x)=∫f(x)dx=F(x)+C{P}{\left({x}\right)}=\int{f{{\left({x}\right)}}}{d}{x}={F}{\left({x}\right)}+{C}.

Since P(a)=0{P}{\left({a}\right)}={0} then 0=P(a)=F(a)+C{0}={P}{\left({a}\right)}={F}{\left({a}\right)}+{C} or C=−F(a){C}=-{F}{\left({a}\right)}.

Therefore, P(x)=F(x)−F(a){P}{\left({x}\right)}={F}{\left({x}\right)}-{F}{\left({a}\right)}.

Newton-Leibniz Formula. Area of figure P(x){P}{\left({x}\right)} that lies under curve y=f(x){y}={f{{\left({x}\right)}}}, above x-axis on interval [a,x]{\left[{a},{x}\right]} is P(x)=F(x)−F(a){P}{\left({x}\right)}={F}{\left({x}\right)}-{F}{\left({a}\right)}.

Returning to our initial problem we have that S=F(b)−F(a){S}={F}{\left({b}\right)}-{F}{\left({a}\right)}.

Area of figure that lies under curve y=f(x){y}={f{{\left({x}\right)}}}, above x-axis and between lines x=a{x}={a} and x=b{x}={b} is S=F(b)−F(a){S}={F}{\left({b}\right)}-{F}{\left({a}\right)} where F{F} is antiderivative of f{f{}}.

Example 2. Find area under y=x2{y}={{x}}^{{2}}, above x-axis on interval [0,2]{\left[{0},{2}\right]}.area under parabola

We have that F(x)=∫x2dx=13x3{F}{\left({x}\right)}=\int{{x}}^{{2}}{d}{x}=\frac{{1}}{{3}}{{x}}^{{3}} (note that we need particular derivative here).

Now, according to Newton-Leibniz Formula required area is S=F(2)−F(0)=13⋅23−13⋅03=83{S}={F}{\left({2}\right)}-{F}{\left({0}\right)}=\frac{{1}}{{3}}\cdot{{2}}^{{3}}-\frac{{1}}{{3}}\cdot{{0}}^{{3}}=\frac{{8}}{{3}}.

All from above can be easily extended on case when function can take negative values. For this we only need to consider areas under x-axis negative.

Indeed, if we want to find area enclose by parabola y=−x2{y}=-{{x}}^{{2}}, x-axis on interval [0,2]{\left[{0},{2}\right]} then according to Newton-Leibniz Formula we have that S=F(2)−F(0){S}={F}{\left({2}\right)}-{F}{\left({0}\right)} where F(x)=∫−x2dx=−13x3{F}{\left({x}\right)}=\int-{{x}}^{{2}}{d}{x}=-\frac{{1}}{{3}}{{x}}^{{3}}. So, S=−83{S}=-\frac{{8}}{{3}}. Since −x2-{{x}}^{{2}} is reflection of x2{{x}}^{{2}} about x-axis then this area lies below x-axis, and, therefore, negative.