definite integral that gives area of a region

I am trying to set up and evaluate the definite integral that gives the area of the region bounded by the graph of the function $\displaystyle y = x ^3 -2x$ and the tangent line to the graph at the point ( -1, 1)

Now, i am aware this is an area problem so i probably have

$\displaystyle a = \int f(x) - g(x) dx$

would g(x) = x ^3 -2x ?

for the tangent line would it be in the form y = mx + b?

How would i find that? I have graphed the x ^3 -2x and made the point ( -1, 1) and can see the region for the area.

Would i need to take the dirivitive of x ^3 -2x and plug in -1 for x?

$\displaystyle x ^3 -2x = 3(-1)^2 - 2 = 1 $

Re: definite integral that gives area of a region

Is the region in question $\displaystyle y = x ^3 -2x$, the tangent line and the X-axis or the Y-axis?

Re: definite integral that gives area of a region

it just says set up and evaluate the definite integral that gives the area of the region bounded by the graph of the function y = x^3 -2x and the tangent line to the graph at the point ( -1, 1)

Re: definite integral that gives area of a region

Hmmm, well it makes a difference in how you set up the integral and the limits of the integral. But going back to your general question, yes, you would take the derivative and then plug in -1 into the derivative. This value then is your slope of the tangent line and since you have the point (-1,1) you can then use y=mx+b to get the equation of the tangent line. Whether it is the X-axis or the Y-axis, f(x) would be the tangent line and g(x) would be $\displaystyle y=x^3 - 2x$ if you decide to do the integral in terms of the variable x. One thing though, if it is the X-axis, you will have two integrals as the g(x) is the x-axis from $\displaystyle -2$ to -$\displaystyle \sqrt{2}$ and then $\displaystyle y=x^3 - 2x$ from -$\displaystyle \sqrt{2}$ to $\displaystyle -1$.

Re: definite integral that gives area of a region

icelated,

Have you looked at the graph of this?

In your Original Post, you found that f '(-1) = 1.

What is the equation of the line that passes through the point (-1,1) and has a slope of 1 ?

Re: definite integral that gives area of a region

The line tangent to $\displaystyle x^3- 2x$ at (-1, 1) crosses the graph of $\displaystyle y= x^3- 2x$ at a second point enclosing a bounded region. What is that other point?

Re: definite integral that gives area of a region

@**HallsofIvy - I dont know!**

if f '(-1) = 1. then, the point (-1, 1) and y = mx + b should be

$\displaystyle y = x + 1$

would that be the f(x) for my other equation?

The graph would be a straight line up one and over one!

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Re: definite integral that gives area of a region

Yes, y = x + 1 is the equation of the tangent line.

Here is a graph of g(x) and the tangent line, done using WolframAlpha.

Added in **Edit**:

As **Youkla** points out below, the equation of the tangent line is y = x + 2 .

Re: definite integral that gives area of a region

Quote:

Originally Posted by

**SammyS** Yes, y = x + 1 is the equation of the tangent line.

Here is a graph of g(x) and the tangent line, done using WolframAlpha.

SammyS, you have a typo. The tangent line is y = x + 2, not x + 1. You state y = x + 2 in the picture correctly.

Re: definite integral that gives area of a region

Thank you guys. I think i have f(x) and g(x) now and can integrate the area. The tangent line is y = x + 2

Re: definite integral that gives area of a region

If i needed to evaluate that would it be between -1 and 2?

Or, am i looking at -1 and 4 by looking at the graph?

Re: definite integral that gives area of a region

Quote:

Originally Posted by

**icelated** If i needed to evaluate that would it be between -1 and 2?

Or, am i looking at -1 and 4 by looking at the graph?

If you integrate over x, (which is the natural way to do it) then use -1 to 2.

If you can find an inverse function, so you ac integrate with respect to y, then use -1 to 4 .

Re: definite integral that gives area of a region

Thats what i thought. Thank you so much!