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

Answer 1

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

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 #

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Emma Aldridge

To compute the value of ∫x^5 dx over the interval [1, 2], you need to integrate the function x^5 with respect to x from 1 to 2. The integral of x^5 with respect to x is (1/6)x^6. Evaluating this from 1 to 2 gives [(1/6)(2^6)] - [(1/6)(1^6)]. Simplifying, you get (64/6) - (1/6) = 63/6 = 10.5. Therefore, the value of the integral over the interval [1, 2] is 10.5.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.