Thread: Find the area between the curves

1. Find the area between the curves

The region R in the first quadrant that is bounded by the graphs of y=x^3, y=4x, and y=-2x+3=0 and that lies below both straight lines. Find the area of this region.

I tried to find the area of the bigger triangle-the smaller region:

I took the integral of -6x+3 from 0 to 1.5 and got -2.25
then i took the integral of -x^3-2x+3 and got .984375

Altogether I get -3.234375. Is this right?

2. Take the integral of these two functions: x^3 and 4x from 0 to the point where 3 - 2x and 4x intersect (1/2). Subtract your answers.

Do the same from 1/2 to the point where 3 - 2x = x^3 (1)

$\displaystyle \int_{0}^{1/2} (4x - x^3)dx + \int_{1/2}^{1} ((3 - 2x) - x^3)dx$

$\displaystyle \int_{0}^{1/2} (4x - x^3)dx + \int_{1/2}^{1} ((3 - 2x) - x^3)dx$

$\displaystyle [2x^2 - \frac {x^4}{4}] + [3x - x^2 - \frac{x^4}{4}]$

$\displaystyle (2(1/2)^2 - \frac {(1/2)^4}{4}) - (2(0)^2 - \frac {(0)^4}{4}) = 1/2 - 1/64$

$\displaystyle (3(1) - (1)^2 - \frac{(1)^4}{4}) - (3(1/2) - (1/2)^2 - \frac{(1/2)^4}{4}) = 2 - 1/4 - 3/2 + 1/4 + 1/64$

$\displaystyle 1/2 - 1/64 + 2 - 1/4 - 3/2 + 1/4 + 1/64 = 1/2 + 2 - 3/2 = 1$