Home > Calculus > Definite and indefinite integrals

How do you evaluate the definite integral #int (x-x^3)dx# from [0,1]?

Answer 1

Isaiah Allen

#1/4#

#int_0^1(x-x^3)dx = int_0^1(x)dx - int_0^1(x^3)dx#

#= x^2/2|_0^1# # - x^4/4|_0^1#

#= (1/2 - 0) - (1/4 - 0)#

#=1/4#

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

Answer 2

Aurora Alvarado

The integral has value #1/4#.

We have by basic integration:

#I = int_0^1 xdx - int_0^1x^3dx#

#I = [1/2x]_0^1 - [1/4x^3]_0^1#

Now find the value using the 2nd fundamental theorem of calculus.

#I = 1/2(1) - 1/2(0) - (1/4(1) - 1/4(0))#

#I = 1/2 - 1/4#

#I = 1/4#

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

Answer 3

Paisley Johnson

To evaluate the definite integral ∫(x - x^3) dx from 0 to 1, you integrate the given function with respect to x and then evaluate it at the upper and lower limits of integration, which are 0 and 1, respectively.

∫(x - x^3) dx = [x^2/2 - x^4/4] evaluated from 0 to 1

Plugging in the upper limit (1): = (1^2/2 - 1^4/4)

Plugging in the lower limit (0):

(0^2/2 - 0^4/4)

Subtract the result of the lower limit from the upper limit: = (1/2 - 1/4) - (0 - 0) = 1/2 - 1/4 = 1/4

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.