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

#### karush

##### Well-known member
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
$$\displaystyle \mathrm{erf(z)}$$ is so called error function.

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

#### karush

##### Well-known member
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
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
Karush, you are making this much harder than it should be!

You are given that $$y'= \int_0^x e^{-t^2}dt$$.

Since $$e^x$$ is positive for all x certainly both $$e^{-t^2}$$ and $$\int_0^x e^{-t^2}dt$$ 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
Karush, you are making this much harder than it should be!

You are given that $$y'= \int_0^x e^{-t^2}dt$$.

Since $$e^x$$ is positive for all x certainly both $$e^{-t^2}$$ and $$\int_0^x e^{-t^2}dt$$ 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

$g’’ = e^{-x^2} > 0$ 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$$Which of the following must be true on the interval$0<x<2$a. g is increasing, and the graph of g is concave up. b. g is increasing, and the graph of g is concave down. c. g is decreasing, and the graph of g is concave up. d. g is decreasing, and the graph of g is concave e. g is decreasing, and the graph of g has a point of inflection on$0<x<2$ok not sure which of options fit (e) looks plausible #### Country Boy ##### Well-known member MHB Math Helper You are not sure? Haven't you read the responses here? You are given that$g'(x)= \int_0^x e^{-t^2}dt$and you have been told that this integral is positive for all x. (Because the integrand is positive for all t.) If a function has positive derivative is it increasing or decreasing? By the "fundamental theorem of Calculus" $$g''= e^{-t^2}$$ which is always positive. If a function has positive second derivative is it convex upward or downward? #### karush ##### Well-known member 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" $$g''= e^{-t^2}$$ 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