For an increasing or decreasing function, the difference between the upper and lower estimate is given by f(b)-f(a)deltaT

Ex¶

suppose \(v(t)=\sqrt{4-t^2}\) on \([0,2]\), \(t\) in seconds, \(v(t)\) in ft/sec. We want to use upper and lower estimates to estimate the total distance traveled from \(t=0\) to \(t=2\)

How small should \(\Delta t\) be so the difference between upper and lower estimates is 0.01?

\((v(2)-v(0))*\Delta t=0.01\) \((0-2)*\Delta t=0.01\) \(2\Delta t=0.01\) \(\Delta t=0.005\)

How many rectangles n should we use? \(\Delta t=\frac{b-a}{n}\) \(0.005=\frac{2}{n}\) \(n=\frac{2}{0.005}\) \(n=400\)

Ex 26¶

A particle moves with velocity \(v(t)=3-t\) along x-axis with time \(t\) in seconds, \(v(t)\) in cm/s for \(0\le t\le 4\).

a) use \(\Delta t=1\) to give lower and upper estimates for the total displacement of the particle. What is the average of those estimates?

Hokora's work: Lower

\(3+2+1+0=6\)

Upper

\(2+1+0+(-1)=2\)

\(\frac{(6+2)}{2}=\fbox{4}\)

b) graph \(v(t)\). Give total displacement. Total displacement: \(|3-(-1)|=\fbox{4}\)

5.2 The Definite Integral¶

divide \([a,b]\) into \(n\) equal pieces. \(v=f(t)\) is velocity function

Want to find change in position of particle from \(t=a\) to \(t=b\). \(\(\Delta t=\frac{b-a}{n}\)\) Left hand sum for change in position

Left Hand Riemann Sum¶

Def: suppose \(f(t)\) is continuous for \(a\le t\le b\). The Definite integral of \(f\) from \(a\) to \(b\), written \(\(\int^{b}_{a}{f(t)dt}\)\) is the limit of the left-hand or right-hand Riemann sums with \(n\) subdivisions of \([a,b]\) as \(n\) gets arbitrarily large (aka \(\Delta t \rightarrow 0\)).

Left-hand Riemann sum¶

\(\(\int^{b}_{a}{f(t)dt}=\lim_{n\rightarrow\infty}\sum^{n-1}_{i=0}{f(t_i)\Delta t}\)\)¶

Right-hand Riemann sum¶

\(\(\int^{b}_{a}{f(t)dt}=\lim_{n\rightarrow\infty}\sum^{n}_{i=1}{f(t_i)\Delta t}\)\)¶

\(f\) is call the integrand and \(a\) and \(b\) called the bounds of integration.

Riemann sums approximate area:¶

When \(f(x)\le 0\) and \(a<b\) Area under graph of \(f\) and above x-axis between \(a\) and \(b\) is: \(\int_{a}^{b}{f(t)dt}\)

When \(f(x)\) is positive for some x-values and negative for others, and \(a<b\): \(\int_{a}^{b}{f(t)dt}\) is the sum of areas above x-axis, minus areas below x-axis.

Ex¶

\(y=sin(x^2)\) \(\int^{\sqrt{2\pi}}_{0}{\sin{(x^2)}dx}\)

Ex 57¶

a) Draw rectangles that give left-sum approximation of \(\int_{0}^{\pi}{\sin{(x)}dx}\) with \(n=2\) (\(n\) is number of rectangles)

b) Do the same, but for \(\int_{0}^{-\pi}{\sin{(x)}dx}\)

c) From (a) and (b), what is the value of the left-hand sum approximation to \(\int_{-\pi}^{\pi}{\sin{(x)}dx}\) with \(n=4\)?

General Riemann sum for \(f\) on \([a,b]\) is sum of form \(\sum_{i=1}^{n}{f(c_i)\Delta t_i}\)

Where \(a=t_0<t_1<t_2<\dots <t_n=b\) for \(i=1,2,\dots ,n\) \(\Delta t_i=t_i-t_{i-1}\) and \(t_{i-1}<=c_i<=t_i\) \(\sum_{i=1}^{n}{(\text{value of }f(t)\text{ at some point in }i\text{th subinterval})*(\text{length of }i\text{th subdivision})}\)

5.3 The Fundamental Theorem of Calculus and Interpretations¶

Velocity \(v=f(t)\) we use small distances in time to estimate change in position, and found that area under curve above x-axis minus area below x-axis bounded by the curve = change in position = definite integral Let \(F(t)\) denote the position function. Then \(F'(t)=f(t)\) and \(\int_{a}^{b}{f(t)dt}\) = Area above minus area below = change in position (\(F(a)-F(b)\))

The Fundamental Theorem of Calculus¶

If \(f\) is continuous on interval \([a,b]\) and \(F'(t)=f(t)\) then \(\int_{a}^{b}{f(t)dt}=F(b)-F(a)\)