Question

Find an equation of the tangent line to the curve
F(x) = integral from 1 to x, e^(-x^2dt) at the point x = 1.

Answer 1

Integral would give f(x)=((e^-(x^2))/2x)-(e^-1)/2
derivative of this is -e^-(x^2)-(e^-(x^2))/2x
put x=1
y=-(3e^-1)/2

Answer 2

Am i right????

Answer 3

sahil has given the slope of the line
if you want the equation of the actual tangent line you need to use point-slope form

but first I would like to be sure that @fero4u123 understood what you did sahil

Answer 4

give me a sec, I'm actually writing it out to see

Answer 5

one the slope of a line on a graph if \(y=f(x)\) at a point \(x=a\) is given by\[y-f(a)=f'(a)(x-a)\]sahil used the fundamental theorem of calculus to find \(f'(1)\)

Answer 6

equation
y=-3e^-1(x-1)/2
am i right????

Answer 7

oh yeah that's right
but the important thing is not the answer, but the understanding

Answer 8

I got it through point slope form as y(0)=0
y-0=m(x-1)

Answer 9

*y(1)=0
but yeah...

Answer 10

did you completely evaluate the integral ? I was under the assumption I could plug in 1 for x and an integral from 1 to 1 is just 0 so that lets me know my point is at
(1,0).
After that I usually take the derivative of F(x) which to me F(x) = e^(-x^2) by the FTC. then F'(x) = e^(-x^2) * 2x
F'(1) = (2/e)
slope of line would be y-0=(2/e)(x-1)

so all and all.. I guess I don't understand because I got a different answer ?

Answer 11

oh yeah!!!!

Answer 12

sry 4 dat!!!!

Answer 13

but \(F(x)\neq e^{-x^2}\)

Answer 14

we don't know F(x)

Answer 15

if the integral was being evaluated to x^2 then you would be right

Answer 16

we don't need to apply the chain rule here

Answer 17

So I do have to evaluate the integral completely to find F(x) ? and I still plug in x=1 for the integral right ?

Answer 18

no, just use FTC\[\frac d{dx}\int_a^xf(t)dt=f(x)\]at least this is the useful form for us in this case

Answer 19

if you doubt where this comes from, look at it this way:\[\frac d{dx}\int_a^xf(t)dt=\frac d{dx}[F(x)-F(a)]=F'(x)-\cancel{F'(a)}^0=F'(x)\]and by assumption of the integral we know that\[F'(x)=f(x)\]therefor\[\frac d{dx}\int_a^xf(t)dt=f(x)\]

Answer 20

note that the derivative of \(F(a)\) is zero, because \(F(a)\) is a constant if a is constant

Answer 21

I'm still confused. I don't see how to get to this "f(x)=((e^-(x^2))/2x)-(e^-1)/2" from the equation.

Answer 22

let me try to draw out what I'm doing so you can see where I'm going wrong.