TITLE: two points and an equation
QUESTION [0 upvotes]: So I think this is the question, I forget how its worded exactly, so bear with me. Suppose we have an equation $5x^3 + bx^2 + cx + d$ cirve that passes through the points $(0,0)$ and $(2,0)$ and the question asks me to solve for $b$ in the equation. How do I do it?

REPLY [1 votes]: You don't have enough information for a unique solution
What you can do is:
$y = 5x^3 + bx^2+ cx + d$
Plug 0 in for x and 0 for y.
$d = 0$
Then plug 2 in for x and 0 for y
$40 + 4b + 2c  = 0$
$b = -10 - \frac 12 c$
And that is as far as you can go.

REPLY [1 votes]: For the point $(0, 0)$, you get
$$0 = 5(0)^3+b(0)^2+c(0)+d \iff d = 0$$
So the equation is essentially $f(x) = 5x^3+bx^2+cx$.
For the point $(2, 0)$, you get
$$0 = 5(2)^3+b(2)^2+c(2) \iff 0 = 40+4b+2c \iff -40 = 4b+2c \iff -20 = 2b+c$$
There isn’t enough information to solve for $b$. However, you can come up with a relation between $b$ and $c$, as shown.


TITLE: $\int _0^1\:2xe^{x^2-1}dx$
QUESTION [0 upvotes]: for the answer I'm getting:
$1+\frac{1}{e}$
but according to symbolab, I get:
$1-\frac{1}{e}$
steps I'm currently taking with u substitution; I get:
$\int \:e^u$
then plugging back u;
$e^{x^2-1}$ (from 0 to 1)
then I compute,
$e^{-1}+e^0$ = $1+\frac{1}{e}$
I'm not sure

REPLY [0 votes]: For the definite integral it would be $e^{1-1} - e^{0-1}$