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How do you compute the value of #int x^5 dx# of #[1,2]#?

1 Answer

Nov 22, 2016

# int_1^2 x^5dx = 21/2 #

Explanation:

Using the power rule for integration; #int x^n=x^(n+1)/(n+1) (+c)#

We have:
# int_1^2 x^5dx = [x^6/6]_1^2 #

# :. int_1^2 x^5dx = 1/6 [x^6]_1^2 #

# :. int_1^2 x^5dx = 1/6 { 2^6 - 1^6 } #
# :. int_1^2 x^5dx = 1/6 { 64- 1 } #
# :. int_1^2 x^5dx = 63/6 #
# :. int_1^2 x^5dx = 21/2 #

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