Find the area of the region bounded by

Answered question

2022-04-23

Find the area of the region bounded by the graphs of the given equations. 𝑦 = 𝑥 2 − 3𝑥 + 3, 𝑦 = −𝑥 + 3

Answer & Explanation

nick1337

nick1337

Expert2022-07-31Added 777 answers

y=x2-3x+3 , y=-x+3

Solve by substitution to find the intersection between the curves.

Eliminate the equal sides of each equation and combine.

x2-3x+3=-x+3

Solve x2-3x+3=-x+3 for x.

x=0,2

Evaluate y when x=0.

y=3

Evaluate y when x=2.

y=1

The solution to the system is the complete set of ordered pairs that are valid solutions.

(0,3)

(2,1)

The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region. The regions are determined by the intersection points of the curves. This can be done algebraically or graphically.

Area=02-x+3dx-02x2-3x+3dx

Integrate to find the area between 0 and 2.

Combine the integrals into a single integral.

02-x+3-(x2-3x+3)dx

Simplify each term.

02-x+3-x2+3x-3dx

Simplify by adding terms.

02-x2+2xdx

Split the single integral into multiple integrals.

02-x2dx+022xdx

Since -1 is constant with respect to x, move -1 out of the integral.

-02x2dx+022xdx

By the Power Rule, the integral of x2 with respect to x is 13x3.

-(13x3]02)+022xdx

Combine 13 and x3.

-(x33]02)+022xdx

Since 2 is constant with respect to x, move 2 out of the integral.

-(x33]02)+202xdx

By the Power Rule, the integral of x with respect to x is 12x2.

-(x33]02)+2(12x2]02)

Simplify the answer.

Combine 12 and x2.

-(x33]02)+2(x22]02)

Substitute and simplify.

Evaluate x33 at 2 and at 0.

-((233)-033)+2(x22]02)

Evaluate x22 at 2 and at 0.

-(233-033)+2(222-022)

Simplify.

43

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