Question

Use Reimann's limit process to find the area of the region between the graph of the function and the x-axis over the indicated interval. y=x^2-x^3 Interval: [-1, 1]

Answer 1

Integrate (x^2-x^3,x,-1,1)

Answer 2

(x^3/3 - x^4/4) evaluated from -1, to 1

Answer 3

-1/3 - 1/4 - 1/3 + 1/4 = -1/3-1/3=-2/3.. i guess more of the area is below the y axis.

Answer 4

We need Delta x, and a way to represent xi in terms of Delta x\[\Delta x=\frac2n,x_i^*=-1+\frac{2i}n\]The Reimann sum is\[\sum_{i=1}^{n}f(x_i^*)\Delta x=\sum_{i=1}^{n}f(\frac{2i}n-1)\cdot\frac2n\]\[=\sum_{i=1}^{n}[(\frac{2i}n-1)^2-(\frac{2i}n-1)^3]\cdot\frac2n\]\[=\sum_{i=1}^{n}(\frac{-8i^3}{n^3}+\frac{16i^2}{n^2}-\frac{10i}n+2)\frac2n\]\[=-\frac{16}{n^4}\sum_{i=1}^{n}i^3+\frac{32}{n^3}\sum_{i=1}^{n}i^2-\frac{20}{n^2}\sum_{i=1}^{n}i+\frac1n\sum_{i=1}^{n}4\]Using the summation formulas for the sum of the first n integers, squares, and cubes we get\[=-\frac{16}{n^4}\frac{n^2(n+1)^2}4+\frac{32}{n^3}\frac{n(n+1)(2n+1)}6-\frac{20}{n^2}\frac{n(n+1)}2+4\]\[=-4\frac{(n+1)^2}{n^2}+\frac{16}3\frac{(n+1)(2n+1)}{n^2}-10\frac{n+1}n+4\]Taking the limit from n to infinity gives the exact area\[\int_{-1}^{1}x^2-x^3dx=\lim_{n \rightarrow \infty}\sum_{i=1}^{n}f(x_i^*)\Delta x\]\[=\lim_{n \rightarrow \infty}-4\frac{(n+1)^2}{n^2}+\frac{16}3\frac{(n+1)(2n+1)}{n^2}-10\frac{n+1}n+4\]\[=-4+\frac{32}3-10+4\]\[=\frac23\]