5.4 Area between Curves¶

\(f(x)=-x^2+4x-1\) \(g(x)=x^2-4x+5\)

\(\int_{a}^{b}{f(x)dx}-\int_{a}^{b}{g(x)dx}\) Area between \(f\) and \(g\) for \(a\le x\le b\) : \(\int_{a}^{b}{f(x)-g(x)dx}\) assuming \(f(x)\le g(x)\) for \(a\le x\le b\)

hokora holy shit put your graph here please

in this picture, brown area = \(\int_{a}^{b}{(-x^2+4x-1)-x^2-4x+5}dx\) to find \(a\) and \(b\), set \(f(x)=g(x)\)

\(\cancel{-x^2}+\cancel{4x}-1=x^2-4x+5\) \(+\cancel{x^2}-\cancel{4x}+1+x^2-4x+1\) \(0=2x^2-8x+6\) \(0=2(x^2-4x+3)\) \(0=2(x-3)(x-1)\) \(x=3,1\) \(\text{Area}=\int_{1}^{3}{(-x^2+4x-1)-(x^2-4x+5)}dx\) \(=\int_{1}^{3}{-2x^2+8x-6}dx\)

6.1 Antiderivates Graphically and Numerically¶

If \(F'=f\), we call \(F\) an antiderivative of \(f\).

Ex:¶

Since \(\frac{d}{dx}(x^2)=2x\), we say \(x^2\) is an antiderivative of \(2x\).

\(\frac{d}{dx}(C)=0\) So any \(C\) is an antiderivative of \(0\)

\(\frac{d}{dx}(x^2+6)=2x\) so \(x^2+6\) is an antiderivative of \(2x\)

In fact, \(x^2+C\) is an antiderivative of \(2x\) for any constant \(C\). We say that \(2x\) has a family of antiderivatives.

Recall the FTC: \(\int_{a}^{b}{f(x)dx}=F(b)-F(a)\) (where \(F'=f\)) \(\Rightarrow F(b)=F(a)+\int_{a}^{b}{F'(x)dx}\)

Ex:¶

The graph of $f'F is given below. Sketch the graph of \(f\) given $f(0)=0

hokora holy shit put your graph here please

\(f\) is increasing when \(f'>0\) \(f\) is decreasing when \(f'<0\) \(f\) is concave up when \(f''>0\) when \(f'\) increasing \(f\) is concave down when \(f''<0\) when \(f'\) decreasing

hokora holy shit put your graph here please

\(f(4)=f(3)+\int_{3}^{4}{f'(x)dx}\) \(f(4)=3+\frac{1}{2}(1)(1)=3.5\)

\(f(5)=f(4)+\int_{4}^{5}{f'(x)dx}\) | \(f(5)=f(3)+\int_{3}^{5}{f(x)dx}\) \(=3.5-\frac{1}{2}(1)(1)\) | \(=3+0=3\) \(=3\) | -|- \(f(4)=f(3)+\int_{3}^{4}{f'(x)dx}\) \(=4+\frac{1}{2}=4.5\)

Ex:¶

Suppose \(f'\) is graphed below and \(f(0)=100\). Find local min's/max's, inflections where \(f\) is inc/dec, where \(f\) is concave up/down, and graph \(f\).

hokora holy shit put your graph here please

\(f\) local min: \(f\) switches from dec to inc = \(f'\) switches from negative to positive \(f\) local max: \(f\) switches from inc to dec = \(f'\) switches from positive to negative \(f\) inflection: \(f\) switches concavity = \(f'\) switches between increasing and decreasing

in the case of the graph above: min's : \(x=0,30\) max's : \(x=20\) inflection pts: \(x=10,25\)

inc: \((0,20)\cup(30,35)\) dec: \((20,30)\) concave up: \((0,10)\cup(25,35)\) concave down: \((10,25)\)

\(f(0)=100\) \(f(10)=f(0)+\int_{0}^{10}{f'(x)dx}\) \(=100+\frac{1}{2}(10)(20)= 100+100=200\) \(f(20)=f(10)+\int_{10}^{20}{f'(x)dx}\) \(=200+100=300\) \(f(25)={f(20)}+\int_{20}^{25}{f'(x)dx}\) \(=300-\frac{1}{2}(5)(10)\) \(=300-25=275\) \(f(20)=f(25)+\int_{25}^{30}{f'(x)dx}\) \(=275-25=250\) \(f(35)=f(30)+\int_{30}^{35}{f'(x)dx}\) \(=250+25=275\)

hokora holy shit put your graph here please

6.2 Constructing Antiderivatives Analytically¶

If \(F'(x)=0\) on an interval, then \(F(x)=C\) for some constant \(C\) on that interval.

What is the more general antiderivative of \(f\)? Suppose \(F\) and \(G\) are antiderivatives of \(f\). \(F'=f\) and \(G'=f\) \(G'=F'\) \(\underline{-F'-F'}\) \(G'-F'=0\) \((G-F)'=0\) \(G-F=C\) for some constant \(C\). \(\Rightarrow G(x)=F(x)+C\)

If \(F\) and \(G\) are both antiderivatives of \(f\), then \(G(x)=F(x)+C\) for some constant \(C\).

Therefore, if \(F\) is an antiderivative of \(f\), then, every antiderivative of \(f\) is of form \(F(x)+C\)

Definition¶

We let \(\int f(x)dx=F(x)+C\) the family of all antiderivatives of \(f\), we call this the indefinite integral of \(f\).

\(\int f(x)dx\) : family of functions \(\int_{a}^{b}{f(x)dx}\) : number representing a signed area (signed area meaning area on top minus area on bottom!)

Ex:¶

\(x^2\) is an antiderivative of \(2x\), so \(\int 2xdx = x^2+C\)

If \(k\) is a constant, \(\int kdk=kx+C\)¶

ie: \(\int 2dx=2x+C\Rightarrow(2x+C)'=2+0=2\)

Power functions¶

\(\frac{d}{dx}(x^2)=2x\Rightarrow\frac{d}{dx}(\frac{x^2}{2})=x\) \(\frac{d}{dx}(x^3)=3x^2\Rightarrow\frac{d}{dx}(\frac{x^3}{3})=x\) \(\frac{d}{dx}(x^4)=4x^3\Rightarrow\frac{d}{dx}(\frac{x^4}{4})=x\) \(\vdots=\vdots\Rightarrow\vdots=\vdots\) \(\frac{d}{dx}(x^{n+1})=(n+1)x^n\Rightarrow\frac{d}{dx}(\frac{x^{n+1}}{n+1})=x\)

\(\int x^ndx=\frac{x^{n+1}}{n+1}+C,n\ne-1\) If \(n=-1\), \(x^-1=\frac{1}{x}\) \(\frac{d}{dx}(\ln{x})=\frac{1}{x},x>0\) \(\frac{d}{dx}(\ln{|1|})=\frac{1}{x}\)

\(\int\frac{a}{x}dx=\ln{|x|}+C\)

\(\int e^xdx=e^x+C\)

\(\frac{d}{dx}(\sin{x})=\cos{x}\) \(\frac{d}{dx}(\cos{x})=-\sin{x}\)

\(\int\cos{x}dx=\sin{x}+C\) \(\int\sin{x}dx=-\cos{x}+C\)

\(\int 3x+x^2dx=3(\frac{x^2}{2})+\frac{x^3}{3}+C\) \(=\frac{3}{2}x^2+\frac{x^3}{3}+C\)