# 5.2 The definite integral  (Page 7/16)

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In the following exercises, express the limits as integrals.

$\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\left({x}_{i}^{*}\right)\text{Δ}x$ over $\left[1,3\right]$

$\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\left(5{\left({x}_{i}^{*}\right)}^{2}-3{\left({x}_{i}^{*}\right)}^{3}\right)\text{Δ}x$ over $\left[0,2\right]$

${\int }_{0}^{2}\left(5{x}^{2}-3{x}^{3}\right)dx$

$\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}{\text{sin}}^{2}\left(2\pi {x}_{i}^{*}\right)\text{Δ}x$ over $\left[0,1\right]$

$\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}{\text{cos}}^{2}\left(2\pi {x}_{i}^{*}\right)\text{Δ}x$ over $\left[0,1\right]$

${\int }_{0}^{1}{\text{cos}}^{2}\left(2\pi x\right)dx$

In the following exercises, given L n or R n as indicated, express their limits as $n\to \infty$ as definite integrals, identifying the correct intervals.

${L}_{n}=\frac{1}{n}\sum _{i=1}^{n}\frac{i-1}{n}$

${R}_{n}=\frac{1}{n}\sum _{i=1}^{n}\frac{i}{n}$

${\int }_{0}^{1}xdx$

${L}_{n}=\frac{2}{n}\sum _{i=1}^{n}\left(1+2\frac{i-1}{n}\right)$

${R}_{n}=\frac{3}{n}\sum _{i=1}^{n}\left(3+3\frac{i}{n}\right)$

${\int }_{3}^{6}xdx$

${L}_{n}=\frac{2\pi }{n}\sum _{i=1}^{n}2\pi \frac{i-1}{n}\text{cos}\left(2\pi \frac{i-1}{n}\right)$

${R}_{n}=\frac{1}{n}\sum _{i=1}^{n}\left(1+\frac{i}{n}\right)\text{log}\left({\left(1+\frac{i}{n}\right)}^{2}\right)$

${\int }_{1}^{2}x\phantom{\rule{0.1em}{0ex}}\text{log}\left({x}^{2}\right)dx$

In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x -axis.

$1+2·2+3·3=14$

$1-4+9=6$

$1-2\pi +9=10-2\pi$

In the following exercises, evaluate the integral using area formulas.

${\int }_{0}^{3}\left(3-x\right)dx$

${\int }_{2}^{3}\left(3-x\right)dx$

The integral is the area of the triangle, $\frac{1}{2}$

${\int }_{-3}^{3}\left(3-|x|\right)dx$

${\int }_{0}^{6}\left(3-|x-3|\right)dx$

The integral is the area of the triangle, 9.

${\int }_{-2}^{2}\sqrt{4-{x}^{2}}dx$

${\int }_{1}^{5}\sqrt{4-{\left(x-3\right)}^{2}}dx$

The integral is the area $\frac{1}{2}\pi {r}^{2}=2\pi .$

${\int }_{0}^{12}\sqrt{36-{\left(x-6\right)}^{2}}dx$

${\int }_{-2}^{3}\left(3-|x|\right)dx$

The integral is the area of the “big” triangle less the “missing” triangle, $9-\frac{1}{2}.$

In the following exercises, use averages of values at the left ( L ) and right ( R ) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.

$\left\{\left(0,0\right),\left(2,1\right),\left(4,3\right),\left(5,0\right),\left(6,0\right),\left(8,3\right)\right\}$ over $\left[0,8\right]$

$\left\{\left(0,2\right),\left(1,0\right),\left(3,5\right),\left(5,5\right),\left(6,2\right),\left(8,0\right)\right\}$ over $\left[0,8\right]$

$L=2+0+10+5+4=21,R=0+10+10+2+0=22,\frac{L+R}{2}=21.5$

$\left\{\left(-4,-4\right),\left(-2,0\right),\left(0,-2\right),\left(3,3\right),\left(4,3\right)\right\}$ over $\left[-4,4\right]$

$\left\{\left(-4,0\right),\left(-2,2\right),\left(0,0\right),\left(1,2\right),\left(3,2\right),\left(4,0\right)\right\}$ over $\left[-4,4\right]$

$L=0+4+0+4+2=10,R=4+0+2+4+0=10,\frac{L+R}{2}=10$

Suppose that ${\int }_{0}^{4}f\left(x\right)dx=5$ and ${\int }_{0}^{2}f\left(x\right)dx=-3,$ and ${\int }_{0}^{4}g\left(x\right)dx=-1$ and ${\int }_{0}^{2}g\left(x\right)dx=2.$ In the following exercises, compute the integrals.

${\int }_{0}^{4}\left(f\left(x\right)+g\left(x\right)\right)dx$

${\int }_{2}^{4}\left(f\left(x\right)+g\left(x\right)\right)dx$

${\int }_{2}^{4}f\left(x\right)dx+{\int }_{2}^{4}g\left(x\right)dx=8-3=5$

${\int }_{0}^{2}\left(f\left(x\right)-g\left(x\right)\right)dx$

${\int }_{2}^{4}\left(f\left(x\right)-g\left(x\right)\right)dx$

${\int }_{2}^{4}f\left(x\right)dx-{\int }_{2}^{4}g\left(x\right)dx=8+3=11$

${\int }_{0}^{2}\left(3f\left(x\right)-4g\left(x\right)\right)dx$

${\int }_{2}^{4}\left(4f\left(x\right)-3g\left(x\right)\right)dx$

$4{\int }_{2}^{4}f\left(x\right)dx-3{\int }_{2}^{4}g\left(x\right)dx=32+9=41$

In the following exercises, use the identity ${\int }_{\text{−}A}^{A}f\left(x\right)dx={\int }_{\text{−}A}^{0}f\left(x\right)dx+{\int }_{0}^{A}f\left(x\right)dx$ to compute the integrals.

${\int }_{\text{−}\pi }^{\pi }\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{1+{t}^{2}}dt$ $\text{(}Hint\text{:}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\text{−}t\right)=\text{−}\text{sin}\left(t\right)\text{)}$

${\int }_{\text{−}\sqrt{\pi }}^{\sqrt{\pi }}\frac{t}{1+\text{cos}\phantom{\rule{0.1em}{0ex}}t}dt$

The integrand is odd; the integral is zero.

${\int }_{1}^{3}\left(2-x\right)dx$ ( Hint: Look at the graph of f .)

${\int }_{2}^{4}{\left(x-3\right)}^{3}dx$ ( Hint: Look at the graph of f .)

The integrand is antisymmetric with respect to $x=3.$ The integral is zero.

In the following exercises, given that ${\int }_{0}^{1}xdx=\frac{1}{2},{\int }_{0}^{1}{x}^{2}dx=\frac{1}{3},$ and ${\int }_{0}^{1}{x}^{3}dx=\frac{1}{4},$ compute the integrals.

${\int }_{0}^{1}\left(1+x+{x}^{2}+{x}^{3}\right)dx$

${\int }_{0}^{1}\left(1-x+{x}^{2}-{x}^{3}\right)dx$

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}=\frac{7}{12}$

${\int }_{0}^{1}{\left(1-x\right)}^{2}dx$

${\int }_{0}^{1}{\left(1-2x\right)}^{3}dx$

${\int }_{0}^{1}\left(1-2x+4{x}^{2}-8{x}^{3}\right)dx=1-1+\frac{4}{3}-2=-\frac{2}{3}$

${\int }_{0}^{1}\left(6x-\frac{4}{3}{x}^{2}\right)dx$

${\int }_{0}^{1}\left(7-5{x}^{3}\right)dx$

$7-\frac{5}{4}=\frac{23}{4}$

In the following exercises, use the comparison theorem .

Show that ${\int }_{0}^{3}\left({x}^{2}-6x+9\right)dx\ge 0.$

Show that ${\int }_{-2}^{3}\left(x-3\right)\left(x+2\right)dx\le 0.$

The integrand is negative over $\left[-2,3\right].$

Show that ${\int }_{0}^{1}\sqrt{1+{x}^{3}}dx\le {\int }_{0}^{1}\sqrt{1+{x}^{2}}dx.$

Show that ${\int }_{1}^{2}\sqrt{1+x}dx\le {\int }_{1}^{2}\sqrt{1+{x}^{2}}dx.$

$x\le {x}^{2}$ over $\left[1,2\right],$ so $\sqrt{1+x}\le \sqrt{1+{x}^{2}}$ over $\left[1,2\right].$

Show that ${\int }_{0}^{\pi \text{/}2}\text{sin}\phantom{\rule{0.1em}{0ex}}tdt\ge \frac{\pi }{4}.$ $\text{(}Hint\text{:}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\ge \frac{2t}{\pi }$ over $\left[0,\frac{\pi }{2}\right]\text{)}$

Show that ${\int }_{\text{−}\pi \text{/}4}^{\pi \text{/}4}\text{cos}\phantom{\rule{0.1em}{0ex}}tdt\ge \pi \sqrt{2}\text{/}4.$

$\text{cos}\left(t\right)\ge \frac{\sqrt{2}}{2}.$ Multiply by the length of the interval to get the inequality.

In the following exercises, find the average value f ave of f between a and b , and find a point c , where $f\left(c\right)={f}_{\text{ave}}.$

$f\left(x\right)={x}^{2},a=-1,b=1$

$f\left(x\right)={x}^{5},a=-1,b=1$

${f}_{\text{ave}}=0;c=0$

$f\left(x\right)=\sqrt{4-{x}^{2}},a=0,b=2$

$f\left(x\right)=\left(3-|x|\right),a=-3,b=3$

$\frac{3}{2}$ when $c=±\frac{3}{2}$

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