# Graph : Finding the Area of a Shaded Region

(See attached file for full problem description with image)

The graph below represents the function f(x) = x3 + 2x2 - 5x - 6. Explain how you process the calculation of the shaded region.

## Solution This solution is **FREE** courtesy of BrainMass!

Given our function f(x) = x^3 + 2x^2 - 5x - 6 the shaded region required is the sum of two areas which we will call A (shaded region above the x axis) and B (shaded region below the x axis).

To find A we need to integrate f(x) with respect to x between two limits i.e. from the lowest value of x for region A, to the greatest value of x for region A.

To find these limits we must calculate the values of x at which f(x) cuts across the x axis, and as you know, these are the roots of our cubic equation.

There are a number of ways to find the roots of our equation - the easiest approach in this case seems to be by factorising the cubic.

To factorise f(x) we can use the polynomial factor theorem which tells us that:

if f(a) = 0 then (x-a) is a factor of f(x).

From your graph you might consider the value: a = -1

f(-1) = (-1)^3 + 2(-1)^2 -5(-1) - 6

= 0

This satisfies the condition of the factor theorem so:

=> (x - (-1)) is a factor of f(x).

We can use this to factorise f(x) completely and so find all the roots:

x^3 + 2x^2 - 5x - 6 = (x + 1) (x^2 + x - 6)

= (x + 1) (x + 3) (x - 2)

=> the roots of f(x) are at x = -3, -1, and 2

which seems to agree with the graph.

Now we know the lower and upper limits for A are (-3 to -1) and those for B are

(-1 to 2).

So we're ready to start integrating:

-1 2

Total area = Int f(x) dx - Int f(x) dx

-3 -1

Note the minus sign is because the region B is under the x axis so the integral will be negative (whereas the area we require is positive).

Remember that differentiating (x^n) gives (n x^(n-1)) so:

Int (x^n) dx = (1/(n+1)) x^(n+1) + c

So we should have no problem integrating our cubic term by term using this equation.

First:

-1 -1

Int f(x) dx = Int (x^3 + 2x^2 - 5x - 6) dx

-3 -3

-1

= [ (1/4)x^4 + (2/3)x^3 - (5/2)x^2 -6x ]

-3

= (1/4 - 2/3 - 5/2 + 6) - (81/4 - 18 - 45/2 + 18)

= 16/3

=> A = 16/3

Secondly:

2 2

Int f(x) dx = Int (x^3 + 2x^2 - 5x - 6) dx

-1 -1

2

= [ (1/4)x^4 + (2/3)x^3 - (5/2)x^2 -6x ]

-1

= (4 + 16/3 - 10 - 12) - (1/4 - 2/3 - 5/2 + 6)

= - 63/4

=> B = 63/4

So we have the answer:

Total area = 16/3 - (-63/4)

= 253/12

(= 21.08)

Note that it was necessary to split the calculation into two integrals. If we had just integrated f(x) from -3 to 2, we would have found the difference between the areas of A and B.

So to summarise the process,

1. Decompose the area into intervals that are either completely above or completely below the x axis.

2. Solve the equation, f(x) = 0, to find the required limits of integration.

3. Carry out the integration

4. Plug in the limits

Hope this helps,

Best wishes,

Paul

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Extra notes for the factorisation if needed:

Given that (x + 1) is a factor of x^3 + 2x^2 - 5x - 6

x^3 + 2x^2 - 5x - 6 = (x + 1) (x^2 + ... )

We need to complete the second bracket on the right. We can see the x^2 is needed so that x x^2 gives us the x^3 term on the left, but what's next? We can see that our x^2 term also is multiplied by the 1 of (x+1) giving us 1 x^2. But from the left hand side we can see we need 2x^2. So we need an extra x^2,

x^3 + 2x^2 - 5x - 6 = (x + 1) (x^2 + x + ... )

Which we can get by adding an x. But this also multiplies with 1 of (x+1) to give us x, when what we really need is (-5x) so we need to subtract 6x from our running total:

x^3 + 2x^2 - 5x - 6 = (x + 1) (x^2 + x - 6)

which you can multiply out to see gives us the correct factorisation.

If this is still confusing you can also look up polynomial division. An alternative method of finding the roots would be the general solution for the cubic equation, but this wouldn't be recommended here.