How do you find the definite integral of #int x^3-xdx# from #[0,2]#?

Question

William Abdalla · Accepted Answer

#int_0^2(x^3-x)dx=2# #int_0^2(x^3-x)dx=[x^4/4-x^2/2]_0^2# #=2^4/4-2^2/2-0# #=2#

Christopher Agresta · Answer

2

use power integration formula: #int x^ndx = x^(n+1) / (n+1) + c# Hence: #int (x^3-x) dx = x^4 / 4 - x^2 / 2 |_0^2 = 2^4 / 4 - 2^2 / 2 =2#

Evelyn Agard · Answer

undefined To find the definite integral of $ \int (x^3 - x) \, dx $ from $ x = 0 $ to $ x = 2 $, follow these steps:

1. Integrate each term of the function separately.
   $ \int x^3 \, dx = \frac{x^4}{4} $
   $ \int x \, dx = \frac{x^2}{2} $

2. Substitute the limits of integration into each integrated term.
   For the term $ x^3 $, evaluate $ \frac{x^4}{4} $ at $ x = 2 $ and $ x = 0 $.
   For the term $ x $, evaluate $ \frac{x^2}{2} $ at $ x = 2 $ and $ x = 0 $.

3. Subtract the result of the evaluation at the lower limit from the result at the upper limit.

$$
\text{{Definite integral}} = \left( \frac{2^4}{4} - \frac{2^2}{2} \right) - \left( \frac{0^4}{4} - \frac{0^2}{2} \right)
$$

4. Simplify the expression.

$$
\text{{Definite integral}} = \left( \frac{16}{4} - \frac{4}{2} \right) - \left( \frac{0}{4} - \frac{0}{2} \right) = \left( 4 - 2 \right) - \left( 0 - 0 \right) = 2
$$

Therefore, the definite integral of $ \int (x^3 - x) \, dx $ from $ x = 0 $ to $ x = 2 $ is 2.