Question

Calculate the area, in square units, bounded by f(x) = 3x³ -3x²-x+8 and

g(x) = 2x³ - 10x² + 29x + 8 over the interval [1, 5].

Answer 1

164

Explanation:

Assuming that you are allowed to use graphing tools:

f(x) is below g(x) for interval [1,3]

g(x) is below f(x) for interval [3,5]

Rules of finding area between two graphs:

`\int\limits^a_b upper - lower \, dx`

For interval [1,3]:

`\int\limits^3_1 {g(x) - f(x)} \, dx`

`\int\limits^3_1 (2x^(3) - 10x^(2) +29x+8)-(3x^(3)-3x^(2) -x+8) \, dx`

`\int\limits^3_1 {-x^(3) -7x^(2) +30x} \, dx`

`= [-(1)/(4)x^(4) - (7)/(3)x^(3) + 15x^(2)]^3_1`

`= [-(1)/(4)3^(4) - (7)/(3)3^(3) + 15*3^(2)] - [-(1)/(4)1^(4) - (7)/(3)1^(3) + 15*1^(2)]`

`= -(81)/(4) - (189)/(3) + 135+(1)/(4) + (7)/(3)- 15\\`

`= (118)/(3)`

For interval [3,5]

`\int\limits^5_3 {f(x)-g(x)} \, dx`

`\int\limits^5_3 (3x^(3)-3x^(2) -x+8)-(2x^(3) - 10x^(2) +29x+8) \, dx`

`\int\limits^5_3 \, dx`

`= [(1)/(4)x^(4) +(7)/(3)x^(3) - 15x^(2)]^5_3`

`= [(1)/(4)5^(4) +(7)/(3)5^(3) - 15*5^(2)] -[(1)/(4)3^(4) +(7)/(3)3^(3) - 15*3^(2)]`

`= (625)/(4) +(875)/(3) - 375 -(81)/(4)} -(189)/(3) + 135`

`=(374)/(3)`

Total area

`= (118)/(3) + (374)/(3)\\`

`= 164`