Consider the area between the graphs x+y=16 and x+4= (y^2). This area can be computed in two different ways using integrals.     First of all it can be computed as a sum of two...

Consider the area between the graphs x+y=16 and x+4= (y^2). This area can be computed in two different ways using integrals.

First of all it can be computed as a sum of two integrals integrate from a to b of f(x)dx + integrate from b to c of g(x)dx

What is the value of a, b, c and what are f(x) and g(x) equal to?

Alternatively this area can be computed as a single integral

integrate from alpha to beta of h(y)dy

Alpha=?, Beta=?, h(y)=?

Either way we find that the area is: ?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should write the equations of both curves in terms of y such that:

`x = 16 - y`

`x = y^2 - 4`

You need to find the limits of integration solving the system of equations such that:

`{(x = 16 - y) , (x = y^2 - 4):}` => `16 - y = y^2 - 4 => y^2 - 4 + y - 16 = 0`

`y^2 + y - 20 = 0`

You should use quadratic formula such that:

`y_(1,2) = (-1+-sqrt(1+80))/2 => y_(1,2) = (-1+-sqrt81)/2`

`y_(1,2) = (-1+-9)/2 => y_1 = 4 ; y_2 = -5`

You need to find the area between the given curves, hence, you need to evaluate the definite integral, with respect to y such that:

`int_-5^4 (y^2 + y - 20) dy`

Using the property of linearity of integrals yields:

`int_-5^4 (y^2 + y - 20) dy = int_-5^4 (y^2)dy +int_-5^4 (y) dy- int_-5^4 (20) dy `

`int_-5^4 (y^2 + y - 20) dy = (y^3/3 + y^2/2 - 20y)|_(-5)^4`

`int_-5^4 (y^2 + y - 20) dy = 64/3 + 125/3 + 16/2 - 25/2 - 80 + 100`

`int_-5^4 (y^2 + y - 20) dy = 189/3 - 9/2 + 20`

`int_-5^4 (y^2 + y - 20) dy = 63 + 20 - 9/2 = 157/2`

Hence, evaluating the area between the given curves, under the given conditions, yields `int_-5^4 (y^2 + y - 20) dy = 63 + 20 - 9/2 = 157/2.`