Find the area of the region bounded by the curves y^2=x, y-4=x, y=-2 and y=1

(Hint: You'll definitely have to sketch this one on paper first.) You get:
a.) 27/2
b.) 22/3
c.) 33/2
d.) 34/3
e.) 14

find the definite integral y^2 - y + 4 evaluated from -2 to 1. I go 16.5 sq units Choice c

33/2 is correct.

Thank you Nade for making this correction! I went back and graphed the functions with pencil to see if I could figure a rough estimate for the area. I found that these function make an odd trapezoid shape. The area of the solid trapezoid portion of the graph is 15 and then there is a small portion on the other side of the y-axis. So, I, too, figured that 33/2 would be the most reasonable answer. Thank you both!

Oh boy, solving math problems is already a circus act, but let's dive in and figure this out together!

First, let's sketch the region on paper. *Drawing sound effects* We have two parabolas, a vertical line, and two horizontal lines. So, it's like we're juggling all these shapes in the air. Quite the circus, isn't it?

Alright, now let's break it down step by step. We need to find the points where these curves intersect to find the boundaries of our region.

The curves y^2 = x and y - 4 = x intersect at y^2 = y - 4, which can be rewritten as y^2 - y + 4 = 0. But solving this equation is no laughing matter. So, let's use the quadratic formula to find the y-coordinate of the intersection points.

*yawn* Oh, sorry. I was having a circus-themed dream about quadratic formulas. Let's get back to the problem.

Using the quadratic formula, we find that the y-coordinate of the intersection points is y = (1 ± √17)/2. Phew! Not the most amusing quadratic equation, but hey, at least we solved it!

Now, let's find the x-coordinates of these intersection points. Substituting the y-coordinate values into the equation y - 4 = x, we get x = (1 ± √17)/2 - 4.

So, we have four points of intersection: ( (1 + √17)/2 - 4, (1 + √17)/2), ( (1 - √17)/2 - 4, (1 - √17)/2), ( (1 + √17)/2 - 4, -2), and ( (1 - √17)/2 - 4, -2).

Now, let's find the area of the bounded region. We integrate the equation y^2 = x from the x-coordinate of the first intersection point to the x-coordinate of the second intersection point.

Oh, this is starting to feel like those tightrope-walking circus clowns balancing on a thin line!

Doing the integration, we find that the area is (1/3) * ∫(1 + √17)/2 - 4 to (1 - √17)/2 - 4 (y^2 - y + 4)dx.

Evaluating this integral gives us the area, which is... *drumroll please*... option d.) 34/3!

Hooray! We made it through the circus of math! I hope my clown act gave you a chuckle while we tackled this problem.

To find the area of the region bounded by the given curves, we first need to sketch the region on paper to get a clear visual representation.

The curves that bound the region are y^2 = x, y - 4 = x, y = -2, and y = 1.

Start by drawing the line y = -2 and y = 1 as horizontal lines.

Next, consider the equation y - 4 = x. Rearranging this equation, we get x = y - 4. This represents a straight line with a slope of 1 and a y-intercept of -4. Draw this line on your graph.

Now let's consider the equation y^2 = x. This equation represents a parabola that opens to the right. Since y^2 is always positive, x has to be positive as well. So we only need to consider the part of the parabola that lies to the right of the y-axis. Sketch the parabola accordingly.

Now we can see that the region we're interested in is the area between these curves. It's a trapezoidal shape with one curved side (the parabola) and one straight side (the line y - 4 = x).

To find the area of this region, we need to integrate the difference between the two curve equations with respect to y.

The equation of the parabola is y^2 = x, which can be rewritten as x = y^2.
The equation of the line is x = y - 4.

To find the limits of integration for y, we need to determine the y-values where the two curves intersect. Setting y - 4 = y^2, we get y^2 - y + 4 = 0. Solving this quadratic equation, we find that it has no real solutions. This means that the curves do not intersect within the given range of -2 to 1 for y.

Therefore, the area of the region bounded by these curves is 0.

None of the answer options provided match this result. It is possible that there was an error in the problem statement or answer choices provided.

Since y^2 = x is not even defined left of the y-axis, the region is

x<3: a triangle bounded by y=x+4, x=-3, y=-2
area = 3*3/2 = 9/2

-3<x<0: a rectangle, area 3*3=9

0<=x<=1: a curved triangular area
area = Int(1-√x)dx [0,1]
= (x - 2/3 x√x)[0,1]
= (1 - 2/3) = 1/3

Total: 9/2 + 9 + 1/3 = 14 - 1/6

Almost (e), but not quite. Typo somewhere?