Question

2 (1o pts) Gven the region bounded hy the graphs x -yo, and y 2 a Sketch a graph of the region. b. Set up the nvo iterated integrals (based upon the rwo choices for order of integration) that represent the area of the region. c Solve either of the two integrals in part b) to find the area of the region -54 3 2-1

Answer 1

(a)

Upper Curve (5,2) y-2 Left Curve Right Curve Area bounded by the curveS 1,0 Lower Curve y-0 x-y 2+1 or y sqrt(x-1)

(b)

The iterative integral would be when we integrate with respect to 'x' to get the area bounded by the curves.

When we are to integrate with respect to 'x' then we need to have function in terms of 'x' that is f(x) or g(x)

$Area,A(x) = \int_{a}^{b}[\textup{Upper Curve } -\textup{Lower Curve}]dx$

Here a = 1 and b = 5

=> $Area,A(x) = \int_{1}^{5}[2 -\sqrt{x-1}]dx$

The other iterative integral would be when we integrate with respect to 'y' to get the area bounded by the curves.

When we are to integrate with respect to 'y' then we need to have function in terms of 'y' that is f(y) or g(y)

$Area,A(y) = \int_{a}^{b}[\textup{Right Curve } -\textup{Left Curve}]dy$

Here a = 0 and b = 2

$Area,A(y) = \int_{0}^{2}[(y^2+1) - (0)]dy$

$Area,A(y) = \int_{0}^{2}(y^2+1)dy$

(c)

Lets solve the second iterative integral that we found in part (b) in order to find the area bounded by the curves.

As this is a bit easier to solve in comparison to the other one.

$Area,A(y) = \int_{0}^{2}(y^2+1) dy$

                        = $(\frac{y^3}{3}+y) (y=0 to 2)$

                        = $(\frac{2^3}{3}+2) - (\frac{0^3}{3}+0)$

                        = $\frac{14}{3}$ units square --------> This is the area bounded by the curves