# 5.2 The definite integral  (Page 8/16)

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$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x,a=0,b=2\pi$

$f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x,a=0,b=2\pi$

${f}_{\text{ave}}=0;c=\frac{\pi }{2},\frac{3\pi }{2}$

In the following exercises, approximate the average value using Riemann sums L 100 and R 100 . How does your answer compare with the exact given answer?

[T] $y=\text{ln}\left(x\right)$ over the interval $\left[1,4\right];$ the exact solution is $\frac{\text{ln}\left(256\right)}{3}-1.$

[T] $y={e}^{x\text{/}2}$ over the interval $\left[0,1\right];$ the exact solution is $2\left(\sqrt{e}-1\right).$

${L}_{100}=1.294,{R}_{100}=1.301;$ the exact average is between these values.

[T] $y=\text{tan}\phantom{\rule{0.1em}{0ex}}x$ over the interval $\left[0,\frac{\pi }{4}\right];$ the exact solution is $\frac{2\phantom{\rule{0.1em}{0ex}}\text{ln}\left(2\right)}{\pi }.$

[T] $y=\frac{x+1}{\sqrt{4-{x}^{2}}}$ over the interval $\left[-1,1\right];$ the exact solution is $\frac{\pi }{6}.$

${L}_{100}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{2}\right)=0.5178,{R}_{100}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{2}\right)=0.5294$

In the following exercises, compute the average value using the left Riemann sums L N for $N=1,10,100.$ How does the accuracy compare with the given exact value?

[T] $y={x}^{2}-4$ over the interval $\left[0,2\right];$ the exact solution is $-\frac{8}{3}.$

[T] $y=x{e}^{{x}^{2}}$ over the interval $\left[0,2\right];$ the exact solution is $\frac{1}{4}\left({e}^{4}-1\right).$

${L}_{1}=0,{L}_{10}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{2}\right)=8.743493,{L}_{100}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{2}\right)=12.861728.$ The exact answer $\approx 26.799,$ so L 100 is not accurate.

[T] $y={\left(\frac{1}{2}\right)}^{x}$ over the interval $\left[0,4\right];$ the exact solution is $\frac{15}{64\phantom{\rule{0.1em}{0ex}}\text{ln}\left(2\right)}.$

[T] $y=x\phantom{\rule{0.1em}{0ex}}\text{sin}\left({x}^{2}\right)$ over the interval $\left[\text{−}\pi ,0\right];$ the exact solution is $\frac{\text{cos}\left({\pi }^{2}\right)-1}{2\pi }.$

${L}_{1}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{\pi }\right)=1.352,{L}_{10}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{\pi }\right)=-0.1837,{L}_{100}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{\pi }\right)=-0.2956.$ The exact answer $\approx -0.303,$ so L 100 is not accurate to first decimal.

Suppose that $A={\int }_{0}^{2\pi }{\text{sin}}^{2}tdt$ and $B={\int }_{0}^{2\pi }{\text{cos}}^{2}tdt.$ Show that $A+B=2\pi$ and $A=B\text{.}$

Suppose that $A={\int }_{\text{−}\pi \text{/}4}^{\pi \text{/}4}{\text{sec}}^{2}tdt=\pi$ and $B={\int }_{\text{−}\pi \text{/}4}^{\pi \text{/}4}{\text{tan}}^{2}tdt.$ Show that $A-B=\frac{\pi }{2}.$

Use ${\text{tan}}^{2}\theta +1={\text{sec}}^{2}\theta .$ Then, $B-A={\int }_{\text{−}\pi \text{/}4}^{\pi \text{/}4}1dx=\frac{\pi }{2}.$

Show that the average value of ${\text{sin}}^{2}t$ over $\left[0,2\pi \right]$ is equal to 1/2 Without further calculation, determine whether the average value of ${\text{sin}}^{2}t$ over $\left[0,\pi \right]$ is also equal to 1/2.

Show that the average value of ${\text{cos}}^{2}t$ over $\left[0,2\pi \right]$ is equal to $1\text{/}2.$ Without further calculation, determine whether the average value of ${\text{cos}}^{2}\left(t\right)$ over $\left[0,\pi \right]$ is also equal to $1\text{/}2.$

${\int }_{0}^{2\pi }{\text{cos}}^{2}tdt=\pi ,$ so divide by the length 2 π of the interval. ${\text{cos}}^{2}t$ has period π , so yes, it is true.

Explain why the graphs of a quadratic function (parabola) $p\left(x\right)$ and a linear function $\ell \left(x\right)$ can intersect in at most two points. Suppose that $p\left(a\right)=\ell \left(a\right)$ and $p\left(b\right)=\ell \left(b\right),$ and that ${\int }_{a}^{b}p\left(t\right)dt>{\int }_{a}^{b}\ell \left(t\right)dt.$ Explain why ${\int }_{c}^{d}p\left(t\right)>{\int }_{c}^{d}\ell \left(t\right)dt$ whenever $a\le c

Suppose that parabola $p\left(x\right)=a{x}^{2}+bx+c$ opens downward $\left(a<0\right)$ and has a vertex of $y=\frac{\text{−}b}{2a}>0.$ For which interval $\left[A,B\right]$ is ${\int }_{A}^{B}\left(a{x}^{2}+bx+c\right)dx$ as large as possible?

The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, $A=\frac{\text{−}b-\sqrt{{b}^{2}-4ac}}{2a}$ and $B=\frac{\text{−}b+\sqrt{{b}^{2}-4ac}}{2a}.$

Suppose $\left[a,b\right]$ can be subdivided into subintervals $a={a}_{0}<{a}_{1}<{a}_{2}<\text{⋯}<{a}_{N}=b$ such that either $f\ge 0$ over $\left[{a}_{i-1},{a}_{i}\right]$ or $f\le 0$ over $\left[{a}_{i-1},{a}_{i}\right].$ Set ${A}_{i}={\int }_{{a}_{i-1}}^{{a}_{i}}f\left(t\right)dt.$

1. Explain why ${\int }_{a}^{b}f\left(t\right)dt={A}_{1}+{A}_{2}+\text{⋯}+{A}_{N}.$
2. Then, explain why $|{\int }_{a}^{b}f\left(t\right)dt|\le {\int }_{a}^{b}|f\left(t\right)|dt.$

Suppose f and g are continuous functions such that ${\int }_{c}^{d}f\left(t\right)dt\le {\int }_{c}^{d}g\left(t\right)dt$ for every subinterval $\left[c,d\right]$ of $\left[a,b\right].$ Explain why $f\left(x\right)\le g\left(x\right)$ for all values of x .

If $f\left({t}_{0}\right)>g\left({t}_{0}\right)$ for some ${t}_{0}\in \left[a,b\right],$ then since $f-g$ is continuous, there is an interval containing t 0 such that $f\left(t\right)>g\left(t\right)$ over the interval $\left[c,d\right],$ and then ${\int }_{d}^{d}f\left(t\right)dt>{\int }_{c}^{d}g\left(t\right)dt$ over this interval.

Suppose the average value of f over $\left[a,b\right]$ is 1 and the average value of f over $\left[b,c\right]$ is 1 where $a Show that the average value of f over $\left[a,c\right]$ is also 1.

Suppose that $\left[a,b\right]$ can be partitioned. taking $a={a}_{0}<{a}_{1}<\text{⋯}<{a}_{N}=b$ such that the average value of f over each subinterval $\left[{a}_{i-1},{a}_{i}\right]=1$ is equal to 1 for each $i=1\text{,…,}\phantom{\rule{0.2em}{0ex}}N.$ Explain why the average value of f over $\left[a,b\right]$ is also equal to 1.

The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, $\begin{array}{c}{\int }_{a}^{b}f\left(t\right)dt={\int }_{{a}_{0}}^{{a}_{1}}f\left(t\right)dt+{\int }_{{a}_{1}}^{{a}_{2}}f\left(t\right)dt+\text{⋯}+{\int }_{{a}_{N+1}}^{{a}_{N}}f\left(t\right)dt={\int }_{{a}_{0}}^{{a}_{1}}1dt+{\int }_{{a}_{1}}^{{a}_{2}}1dt+\text{⋯}+{\int }_{{a}_{N+1}}^{{a}_{N}}1dt\hfill \\ =\left({a}_{1}-{a}_{0}\right)+\left({a}_{2}-{a}_{1}\right)+\text{⋯}+\left({a}_{N}-{a}_{N-1}\right)={a}_{N}-{a}_{0}=b-a.\hfill \end{array}$ Dividing through by $b-a$ gives the desired identity.

Suppose that for each i such that $1\le i\le N$ one has ${\int }_{i-1}^{i}f\left(t\right)dt=i.$ Show that ${\int }_{0}^{N}f\left(t\right)dt=\frac{N\left(N+1\right)}{2}.$

Suppose that for each i such that $1\le i\le N$ one has ${\int }_{i-1}^{i}f\left(t\right)dt={i}^{2}.$ Show that ${\int }_{0}^{N}f\left(t\right)dt=\frac{N\left(N+1\right)\left(2N+1\right)}{6}.$

${\int }_{0}^{N}f\left(t\right)dt=\sum _{i=1}^{N}{\int }_{i-1}^{i}f\left(t\right)dt=\sum _{i=1}^{N}{i}^{2}=\frac{N\left(N+1\right)\left(2N+1\right)}{6}$

[T] Compute the left and right Riemann sums L 10 and R 10 and their average $\frac{{L}_{10}+{R}_{10}}{2}$ for $f\left(t\right)={t}^{2}$ over $\left[0,1\right].$ Given that ${\int }_{0}^{1}{t}^{2}dt=0.\stackrel{–}{33},$ to how many decimal places is $\frac{{L}_{10}+{R}_{10}}{2}$ accurate?

[T] Compute the left and right Riemann sums, L 10 and R 10 , and their average $\frac{{L}_{10}+{R}_{10}}{2}$ for $f\left(t\right)=\left(4-{t}^{2}\right)$ over $\left[1,2\right].$ Given that ${\int }_{1}^{2}\left(4-{t}^{2}\right)dt=1.\stackrel{–}{66},$ to how many decimal places is $\frac{{L}_{10}+{R}_{10}}{2}$ accurate?

${L}_{10}=1.815,{R}_{10}=1.515,\frac{{L}_{10}+{R}_{10}}{2}=1.665,$ so the estimate is accurate to two decimal places.

If ${\int }_{1}^{5}\sqrt{1+{t}^{4}}dt=41.7133...,$ what is ${\int }_{1}^{5}\sqrt{1+{u}^{4}}du?$

Estimate ${\int }_{0}^{1}tdt$ using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value ${\int }_{0}^{1}tdt?$

The average is $1\text{/}2,$ which is equal to the integral in this case.

Estimate ${\int }_{0}^{1}tdt$ by comparison with the area of a single rectangle with height equal to the value of t at the midpoint $t=\frac{1}{2}.$ How does this midpoint estimate compare with the actual value ${\int }_{0}^{1}tdt?$

From the graph of $\text{sin}\left(2\pi x\right)$ shown:

1. Explain why ${\int }_{0}^{1}\text{sin}\left(2\pi t\right)dt=0.$
2. Explain why, in general, ${\int }_{a}^{a+1}\text{sin}\left(2\pi t\right)dt=0$ for any value of a .

a. The graph is antisymmetric with respect to $t=\frac{1}{2}$ over $\left[0,1\right],$ so the average value is zero. b. For any value of a , the graph between $\left[a,a+1\right]$ is a shift of the graph over $\left[0,1\right],$ so the net areas above and below the axis do not change and the average remains zero.

If f is 1-periodic $\left(f\left(t+1\right)=f\left(t\right)\right),$ odd, and integrable over $\left[0,1\right],$ is it always true that ${\int }_{0}^{1}f\left(t\right)dt=0?$

If f is 1-periodic and ${\int }_{0}^{1}f\left(t\right)dt=A,$ is it necessarily true that ${\int }_{a}^{1+a}f\left(t\right)dt=A$ for all A ?

Yes, the integral over any interval of length 1 is the same.

questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The