Hello! I'm having trouble understanding how I'm supposed to work out this problem. Any help would be appreciated!

Question

Find the area of the region bounded by the curve y = f(x) = x3 – 4x + 1 and the tangent line to the curve y = f(x) at (–1,4).

Answer 1

hints:
steps:
1. find the equation of the tangent line.
The slope m of the line equals f'(-1), and passes through (-1,f(-1))=(-1,4)=(x1,y1).
The equation of the line is then
y-y1=m(x-x1), or g(x)=m(x-x1)+y1
Draw a sketch of f(x) and g(x) for your own information.
2. Find intersections of the tangent line with f(x). On of the intersection points is evidently the tangent point, (-1,4), and the other would be denoted (x2,y2).
Then [-1,x2] delimits the region whose area is sought.
3. It would be relatively easy to
evaluate the definite integral
I=∫(g(x)-f(x))dx
for x=-1 to x2.
Since g(x) is always above f(x), the integral gives the area required.

Answer 2

your first task is to find the tangent line at (-1,4). Since that is y = -x+3, you need to find where else the line and the curve intersect. That will be at (2,1).

See the curves at

http://www.wolframalpha.com/input/?i=x%5E3-4x%2B1%3D3-x

So, now just add up all the little strips of width dx, via

∫[-1,2] (3-x)-(x^3-4x+1) dx = 27/4