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[SOLVED] 4.2.251 AP Calculus Exam.....must be true on the interval 0<x<2

karush

Well-known member
Jan 31, 2012
2,678
View attachment 9527

ok from online computer I got this

$\displaystyle\int_0^x e^{-t^2}=\frac{\sqrt{\pi }}{2}\text{erf}\left(t\right)+C$

not sure what erf(t) means
 

Theia

Well-known member
Mar 30, 2016
92
\(\displaystyle \mathrm{erf(z)}\) is so called error function.

But you shouldn't need to know that to be able to solve the problem.
 

karush

Well-known member
Jan 31, 2012
2,678
ok but isn't this just going to be $y=\dfrac{\pi}{2}$ which is just a horizontal line
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655
ok but isn't this just going to be $y=\dfrac{\pi}{2}$ which is just a horizontal line
the integral function defining $g’$ is not a constant

the graph of $e^{-t^2}$ is a bell shaped curve, symmetrical to the y axis and always positive.

for $0<x<2$, $g’ > 0$ (why?)

also, $g’’ > 0$ (why again?)

... what does that say about the behavior of $g$?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Karush, you are making this much harder than it should be!

You are given that [tex]y'= \int_0^x e^{-t^2}dt[/tex].

Since [tex]e^x[/tex] is positive for all x certainly both [tex]e^{-t^2}[/tex] and [tex]\int_0^x e^{-t^2}dt[/tex] are positive for all x and t. The derivative and second derivative of y are positive for all x. That is all you need.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655
Karush, you are making this much harder than it should be!

You are given that [tex]y'= \int_0^x e^{-t^2}dt[/tex].

Since [tex]e^x[/tex] is positive for all x certainly both [tex]e^{-t^2}[/tex] and [tex]\int_0^x e^{-t^2}dt[/tex] are positive for all x and t. The derivative and second derivative of y are positive for all x. That is all you need.
$g’’ = e^{-x^2} > 0$ for all $x$

$g’ > 0$ for $x > 0$, $g’=0$ at $x=0$

$g’ < 0$ for $x < 0$
 

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karush

Well-known member
Jan 31, 2012
2,678
$g’’ = e^{-x^2} > 0$ for all $x$$g’ > 0$ for $x > 0$, $g’=0$ at $x=0$
$g’ < 0$ for $x < 0$
Let g be a function with first derivative given by
$$\displaystyle g'=\int_0^x e^{-t^2}\, dt$$
Which of the following must be true on the interval $0<x<2$
a. g is increasing, and the graph of g is concave up.
b. g is increasing, and the graph of g is concave down.
c. g is decreasing, and the graph of g is concave up.
d. g is decreasing, and the graph of g is concave
e. g is decreasing, and the graph of g has a point of inflection on $0<x<2$

ok not sure which of options fit (e) looks plausible
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
426
You are not sure? Haven't you read the responses here?
You are given that $g'(x)= \int_0^x e^{-t^2}dt$ and you have been told that this integral is positive for all x. (Because the integrand is positive for all t.) If a function has positive derivative is it increasing or decreasing?

By the "fundamental theorem of Calculus" [tex]g''= e^{-t^2}[/tex] which is always positive. If a function has positive second derivative is it convex upward or downward?
 

karush

Well-known member
Jan 31, 2012
2,678
You are not sure? Haven't you read the responses here?
You are given that $g'(x)= \int_0^x e^{-t^2}dt$ and you have been told that this integral is positive for all x. (Because the integrand is positive for all t.) If a function has positive derivative is it increasing or decreasing?

By the "fundamental theorem of Calculus" [tex]g''= e^{-t^2}[/tex] which is always positive. If a function has positive second derivative is it convex upward or downward?
g is bell shaped so it convex down