Given two equation, find intersect point, enclosed area and need to sketch graph!

Question

Determine each of the following.

i. Sketch 𝑦1 and 𝑦2 in the same graph.

ii. Shaded the area of the region bounded.

iii. Find the intersection points.

iv. Find the enclosed area.

Answer 1

Hi Isaac,

Sub-part i and ii

For graphing I suggest you use Desmos Online Graphing Calculator. By graphing these two equations you can visualize the bounded area and note that y₂ = x + 5 is the top bound of the area and y₁ = 2x² + 2 is the bottom bound.

Sub-part iii

To solve for the intersection points of two lines simply set the two equations equal to one another and solve.

2x² + 2 = x + 5

2x² - x - 3 = 0

(2x - 3)(x+1) = 0

The two lines would thus intersect at x = 3/2 and x = -1. Use the graph to verify that this is correct.

Sub-part iv

To solve for the area between two curves use the formula

∫_a^b f(x) - g(x) dx

where f(x) is the top bound of the area, g(x) is the bottom bound of the area, and [a,b] is the interval over which we are solving for the area.

In this case we know f(x) = x + 5, g(x) = 2x² + 2, and [a,b] = [-1, 3/2] so

∫_-1^1.5 x + 5 - (2x² + 2) dx

∫_-1^1.5 x + 5 - 2x² - 2 dx = 5.20833

I hope this helps! If you have any questions let me know.

Best of luck,

Bryce