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How do you find the definite integral of #2x* (x-3)^5 dx# from #[-1, 2]#?

Answer 1

Ezra Adams

# = 4101/7#

we'll use IBP

#int u v' = uv - int u'v#

so for #int_{-1}^2 dx qquad 2x (x-3)^5 #

we can set

#u = 2x, u' = 2# #v' = (x-3)^5, v = 1/6(x-3)^6#

#implies [2x * 1/6(x-3)^6]\_(-1)^(2) - int_{-1}^2 dx qquad 2 * 1/6(x-3)^6 #

# =[1/3 x (x-3)^6]\_(-1)^(2) - 1/3 int_{-1}^2 dx qquad (x-3)^6#

# =[1/3 x (x-3)^6 - 1/3* 1/7(x-3)^7 ]\_(-1)^(2)#

# =[1/3 x (x-3)^6 - 1/21(x-3)^7 ]\_(-1)^(2)#

#= 2/3 + 1/21 + 4^6/3 - 4^7/21#

# = 4101/7#

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Answer 2

Luke Alvarez

To find the definite integral of (2x \cdot (x - 3)^5) with respect to (x) from (-1) to (2), you would first integrate the function with respect to (x) and then evaluate the result at the upper and lower bounds of the interval, and finally subtract the value of the integral at the lower bound from the value at the upper bound.

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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.