How do you evaluate the definite integral $\int 5 x^{3} d x$ $int 5x^3 dx$ from $[- 1, 1]$ $[-1,1]$ ?

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Answer 1 · 2016-09-20T11:31:19

$0$ $0$ .

Using the following well-known Result , we can immediately say,

without any calculation , that, $\int_{- 1}^{1} 5 x^{3} d x = 0$ $int_-1^1 5x^3dx=0$ .

$Result : If f is continuous odd function defined on [- a, a],$ $" Result : If f is continuous odd function defined on "[-a,a],$

$then, \int_{- a}^{a} f (x) d x = 0$ $" then, "int_-a^a f(x)dx=0$ .

However, we work out the Soln. as shown below :-

$I = \int_{- 1}^{1} 5 x^{3} d x = 5 \int_{- 1}^{1} x^{3} d x = 5 {[\frac{x^{4}}{4}]}_{- 1}^{1}$ $I=int_-1^1 5x^3dx=5int_-1^1 x^3dx=5[x^4/4]_-1^1$

$= \frac{5}{4} [1 - {(- 1)}^{4}] = \frac{5}{4} [1 - 1]$ $=5/4[1-(-1)^4]=5/4[1-1]$ .

$:. int_-1^1 5x^3dx=0$ .

How do you evaluate the definite integral int 5x^3 dx∫5x3dxint 5x^3 dx from [-1,1][−1,1][-1,1]?