Find the area of the region between the graphs of

Question

f(x) = 3x³ – x² – 10x and g(x) = -x² + 2x

Answer 1

Find the area of the region between the graphs of. f(x) = 3x³ – x² – 10x and g(x) = -x² + 2x

Detailed Solution:

First, find intersection points:

f(x) = g(x) ==> 3x³ – x² – 10x = -x² + 2x ==> 3x³ – x² – 10x + x² – 2x = 0 ==> 3x³ – 12x = 0 ==>

3x(x² – 4) = 0 ==> 3x = 0 and x² – 4 = 0 ==> x = 0 and x² = 4 ==> x = 0 and x = ±2

Therefore, intersection points are: x = -2 x = 0 x = 2

Sketch the graphs, you will see:

* On interval (a, b) = (-2, 0), f(x) is Top curve and g(x) is Bottom curve

* On interval (b, c) = (0, 2), g(x) is Top curve and f(x) is Bottom curve

Area = ∫_a^b [ Top – Bottom ] + ∫_b^c [ Top – Bottom ]==> Area = ∫_a^b [ f(x) – g(x) ]dx + ∫_b^c [ g(x) – f(x) ]dx==>

Area = ∫_-2⁰ [ (3x³ – x² – 10x) – (-x² + 2x) ]dx + ∫₀² [ (-x² + 2x) – (3x³ – x² – 10x) ]dx ==>

Area = ∫_-2⁰ [ 3x³ – x² – 10x + x² – 2x) ]dx + ∫₀² [ -x² + 2x – 3x³ + x² + 10x) ]dx ==>

Area = ∫_-2⁰ [ 3x³ – 12x ]dx + ∫₀² [ -3x³ + 12x ]dx ==>

Area = [ 3x⁴/4 – 12x²/2 ]|_-2⁰ + [ -3x⁴/4 + 12x²/2 ]|₀² ==>

Area = [ 3x⁴/4 – 6x² ]|_-2⁰ + [ -3x⁴/4 + 6x² ]|₀² ==>

Area = [ (3(0)⁴/4 – 6(0)²) – (3(-2)⁴/4 – 6(-2)²) ] + [ (-3(2)⁴/4 + 6(2)²) – (-3(0)⁴/4 + 6(0)²) ==>

Area = [ 0 – 3(16)/4 + 6(4) ] + [ -3(16)/4 + 6(4) – (0) ==>

Area = [ 0 – 12 + 24 ] + [ -12 + 24 – 0] ==>

Area = 0 – 12 + 24 – 12 + 24 – 0 ==>

Area = 24

Answer 2

Muhammad, I hope you have done this sort of problem before

1) get the limits of integration by finding where f and g intersect (set them equal to each other and solve)

2) integrate (higher - lower) or integrate either way and take the value of the positive integral

Hope this helps--Andrew

PS there is an easy way to get an approximation using your TI-84-- hope you have done that before