- #1

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[tex] \int_{0}^{1} {2x^3 ln (x)} dx [/tex]

My answer was -1/8. Is this correct or not?

- Thread starter LinkMage
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- #1

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[tex] \int_{0}^{1} {2x^3 ln (x)} dx [/tex]

My answer was -1/8. Is this correct or not?

- #2

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I got the integral to be...LinkMage said:

[tex] \int_{0}^{1} {2x^3 ln (x)} dx [/tex]

My answer was -1/8. Is this correct or not?

[tex] [(2x^{4}(lnx-1)-3x^{4}/2)/4] [/tex]

ln 0 is infinate, thus the integral beteween 1 and 0 is infinate?!...

- #3

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Here is what I did (if I remember correctly):

[tex] \int {2x^3ln(x)} dx [/tex]

[tex]u = ln (x) [/tex] [tex] v' = 2x^3dx [/tex]

[tex]u' = \frac {1} {x} [/tex] [tex] v = \frac {1} {2} x^4 [/tex]

[tex]\int {2x^3ln(x)} dx = {\frac {1} {2} x^4} ln (x) - \int {{\frac {1} {x}}{\frac {1} {2} x^4}} dx = {\frac {1} {2} x^4} ln (x) - {\frac {1} {8} x^4} = x^4(\frac {1} {2} ln (x) - \frac {1} {8}) [/tex]

[tex] \int {2x^3ln(x)} dx [/tex]

[tex]u = ln (x) [/tex] [tex] v' = 2x^3dx [/tex]

[tex]u' = \frac {1} {x} [/tex] [tex] v = \frac {1} {2} x^4 [/tex]

[tex]\int {2x^3ln(x)} dx = {\frac {1} {2} x^4} ln (x) - \int {{\frac {1} {x}}{\frac {1} {2} x^4}} dx = {\frac {1} {2} x^4} ln (x) - {\frac {1} {8} x^4} = x^4(\frac {1} {2} ln (x) - \frac {1} {8}) [/tex]

Last edited:

- #4

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My calculator got -1/8, and he is smart And your work looks correct too

- #5

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And then:

[tex] \lim_{\substack{s\rightarrow 0^+}} ${\displaystyle x^4(\frac {1} {2} ln (x) - \frac {1} {8})|_{s}^{1}}$ = \lim_{\substack{s\rightarrow 0^+}} {[1^4(\frac {1} {2} ln (1) - \frac {1} {8})] - {[s^4(\frac {1} {2} ln (s) - \frac {1} {8})] = -\frac {1} {8} - 0 = -\frac {1} {8} [/tex]

[tex] \lim_{\substack{s\rightarrow 0^+}} ${\displaystyle x^4(\frac {1} {2} ln (x) - \frac {1} {8})|_{s}^{1}}$ = \lim_{\substack{s\rightarrow 0^+}} {[1^4(\frac {1} {2} ln (1) - \frac {1} {8})] - {[s^4(\frac {1} {2} ln (s) - \frac {1} {8})] = -\frac {1} {8} - 0 = -\frac {1} {8} [/tex]

Last edited:

- #6

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You have x = 1, which means 1((1/2)ln(1) - 1/8) = -1/8

You have x= 0, which means 0(something) = 0

so the answer is -1/8

- #7

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BTW, what calculator do you have? I want one like that. :P

- #8

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- #9

HallsofIvy

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And they'd be perfectly correct to take marks off if you said:LinkMage said:

BTW, what calculator do you have? I want one like that. :P

"You have x= 0, which means 0(something) = 0"

0(something) is

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