Question

Evaluate the definite integral.
from e to 1

6t^3 ln(t) dt

Answer 1

You must perform integration by parts

Answer 2

i did that and got 3/2e^4ln(e)-3/8 e^4 - 3/2 (1)^4 ln(1)-3/8 (1)^4

Answer 3

ignoring the limits for a sec
what did you if you have the indefinite integral
6t^3*ln(t) dt

Answer 4

i got 3/2 t^4 ln(t)- 3/8 t^4

Answer 5

here u will apply ILATE rule..

Answer 6

ok that looks awesome

Answer 7

then when i plug in i get confuseed

Answer 8

\[(\frac{3}{2} e^4 \ln(e)-\frac{3}{8}(e^4))-(\frac{3}{2}1^4\ln(1)-\frac{3}{8}(1)^4)\]
well you can replace ln(e) with 1
and replace ln(1) with 0

Answer 9

yes i did that still not getting the right answer for my hw idk what I'm doing wrong

Answer 10

\[(\frac{3}{2}e^4-\frac{3}{8}e^4)-(0-\frac{3}{8}) \\ \frac{3}{2}e^4-\frac{3}{8}e^4+\frac{3}{8} \]

Answer 11

And then combine those two fractions that are like terms

Answer 12

you need to multiply that first fraction by 4/4
then add those first two terms

Answer 13

add as in subtract of course because the operation is -

Answer 14

so its 12/8e^4 - 6/8 e^4 ???

Answer 15

i got this
-(3/2) - (3/8)(1-e^4)

Answer 16

why did you put -6/8 instead of -3/8?

Answer 17

oh i added both 3/8 by accident

Answer 18

\[\frac{12}{8}e^4-\frac{3}{8}e^4+\frac{3}{8}=\frac{9}{8}e^4+\frac{3}{8}\]

And you can write that as one fraction if your heart so desires

Answer 19

thank you!!

Answer 20

and you meant from 1 to e right?

Answer 21

i have another question

Answer 22

because we looked at it from 1 to e not from e to 1

Answer 23

yes

Answer 24

ok

Answer 25

can u see it

Answer 26

Looks like another integration by parts

Answer 27

yes but whats the u

Answer 28

you want to differentiate the poly
integrate the exponential

Answer 29

so u=poly and dv=exp

Answer 30

whats poly

Answer 31

x^2 ?

Answer 32

a nickname for polynomial

Answer 33

yea you say 4x^2 or x^2

Answer 34

ok and whats the anti derive of e^-4x

Answer 35

\[\int\limits_{}^{}e ^{ax} dx \text{ \let } u=ax = > du=a dx => \frac{du}{a} =dx \]
\[\int\limits_{}^{}e^u \frac{du}{a}=\frac{1}{a} e^u+C=\frac{1}{a}e^{ax}+c , a \neq 0 \]

Answer 36

\[\int\limits_{}^{}e^{ax} dx=\frac{e^{ax}}{a}+c\]

Answer 37

ok so now its [x^2((e^-4)/-4) -\int\limits e^-4/-4\]

Answer 38

integral***

Answer 39

whats the set up sorry

Answer 40

one sec trying to read

Answer 41

having problems reading that

Answer 42

lemme re write it

Answer 43

4(1/2) x^2(e^-4x/-4) - integral e^-4x

Answer 44

\[\int\limits_{}^{}x^2 e^{-4x} dx \]
\[x^2 \frac{e^{-4x}}{-4}-\int\limits_{}^{}2x \frac{e^{-4x}}{-4} dx\]
So you got here...
I think you should three terms

Answer 45

we will come back later and multiply the 4

Answer 46

ok got it

Answer 47

next step

Answer 48

idk the integral

Answer 49

you must perform integration by parts again

Answer 50

ah ok

Answer 51

whats the u

Answer 52

2x?

Answer 53

the u will remain the poly
yes

Answer 54

whats the anti deriv of e^-4x/-4

Answer 55

well if you don't ignore the -4 already underneath the the integral of e^{-4x)/-4
is actually e^(-4x)/16

Answer 56

?

Answer 57

I thought you were integrating e^(-4x)

Answer 58

I was telling you how to integrate e^(-4x)/-4 which is the same way you just need to take care of the -4 on bottom

Answer 59

the antiderivative of e^(-4x) is e^(-4x)/-4
the antiderivative of e^(-4x)/-4 is e^(-4x)/16

Answer 60

here I will write it in pretty symbols
-1/4 is just a constant multiple

but okay so we have
\[\int\limits_{}^{}e^{-4x} dx =\frac{e^{-4x}}{-4}+C \]
If we multiply both sides by -1/4 we have
\[\frac{-1}{4} \int\limits_{}^{}e^{-4x} dx =\frac{-1}{4}(\frac{e^{-4x}}{-4}+c)\]
\[\int\limits_{}^{}\frac{e^{-4x}}{-4} dx=\frac{e^{-4x}}{16}+K\]