How do you find #\int _ { 0} ^ { 1} x ^ { 5} + x ^ { 3} d x#?

1 Answer
Aug 8, 2017

The integral equals #5/12#

Explanation:

Start by separating the integrals, it will be easier to work with.

#I = int_0^1 x^5dx + int_0^1 x^3dx#

Use the power rule for integration, which states that #int(x^n)dx = x^(n + 1)/(n + 1) + C#.

#I = [1/6x^6]_0^1 + [1/4x^4]_0^1#

Now use the fundamental theorem of calculus to evaluate, which states that #int_a^b f(x) = F(b) - F(a)#, where #F'(x) = f(x)# and #f(x)# is a continuous function on #[a, b]#.

#I = 1/6 - 0 + 1/4 - 0#

#I = 5/12#

Hopefully this helps!