# Thread: Help finding a tangent to integral

1. ## Help finding a tangent to integral

given is F(x)= the integral from x=1 to x=2x^3 of the function (3t-4)^2 dt

I need to find the equation of the tangent line to F(x) at x=-1

so far i have:

f=600 (which is the slope of F at x=-1)

i need help finding the (x,y) coordinates

2. Originally Posted by gamefreeze
given is F(x)= the integral from x=1 to x=2x^3 of the function (3t-4)^2 dt

I need to find the equation of the tangent line to F(x) at x=-1

so far i have:

f=600 (which is the slope of F at x=-1)

i need help finding the (x,y) coordinates
$\displaystyle F(x)=\int_1^{2x^3}(3t-4)^2dt$

Let $\displaystyle u=2x^3$

$\displaystyle F'(x)=\frac{du}{dx}\frac{d}{du}\int_1^u(3t-4)^2dt$

$\displaystyle F'(x)=6x^2[3(2x^3)-4]^2$

$\displaystyle F'(-1)=600$

Ok, In get the same thing there. Think of the equation of the tangent line as follows:

$\displaystyle y-F(-1)=600[x-(-1)]$

$\displaystyle y-F(-1)=600(x+1)$

So you need to find

$\displaystyle F(-1)=\int_1^{2(-1)^3}(3t-4)^4dt$

So that's just a basic definite integral right? So just evaluate F(-1) by using the substitution rule.