Question

Hi I have a question on intergration:

Find the mean values of the following function in the given intervals.

(5x-3)√x, 1≤x≤4

Answer 1

the answer should be 16 am struggling with the numeric part as always.

Answer 2

The mean value is\[\frac{ 1 }{ b-a }\int\limits_{a}^{b}f(x)dx=\frac{ 1 }{ 3 }\int\limits_{1}^{4}(5x-3)\sqrt{x}dx\]You can integrate as follows: expand the integrand and write with exponents:\[\frac{ 1 }{ 3 }\int\limits_{1}^{4}(5x \sqrt{x}-3\sqrt{x})dx=\frac{ 1 }{ 3 }\int\limits_{1}^{4}(5x^{\frac{ 3 }{ 2 }}-3x^{\frac{ 1 }{ 2 }})dx\]

Answer 3

This becomes:\[\frac{ 1 }{ 3 }\left[ 5 \cdot \frac{ 2 }{ 5 }x^{\frac{ 5 }{ 2 }}-3 \cdot \frac{ 2 }{ 3 }x^{\frac{ 3 }{ 2 }} \right]_{1}^{4}\]We have to try to somehow make this look like 16...

Answer 4

am trying to do that...

Answer 5

MAybe we sould try to see what is 4^(5/2) before doing this: \[4^{\frac{ 5 }{ 2 }}=(4^{\frac{ 1 }{ 2 }})^5=2^5=32\]

Answer 6

am gettting 19 as an overall answer

Answer 7

And 4^(3/2)=2^3=8
So now we have:\[\frac{ 1 }{ 3 }\left( (2 \cdot 32-2 \cdot8)-(2-2) \right)=\frac{ 1 }{ 3 }(64-16-0)=\frac{ 48 }{ 3 }=16\]

Answer 8

i made a mistake by writing the 2-3 on the second part and not 2-2 thanks buddy i really appreciate it cheers

Answer 9

Welcome