Question

PLEASE HELP! URGENT! MEDAL! @mathmale

Answer 1

@mathstudent55 @mathmale @ganeshie8 @dan815 @Hero

Answer 2

@mathmath333 @surjithayer

Answer 3

u have to try with substituting \(a=-1,1\) i guess

Answer 4

average value of \(f(x)\) over the interval \([a,b]\) is given by :
\[\dfrac{1}{b-a}\int\limits_a^b f(x)\, dx\]

Answer 5

I think I'm supposed to find it with the variable a, and not substitute anything for it

Answer 6

average value of \(a\sqrt{x}\) over the interval \([0,a]\) is given by :
\[\dfrac{1}{a-0}\int\limits_0^a a\sqrt{x}\, dx\]

Answer 7

should be easy to evaluate

Answer 8

I got \(\Large \frac{2a^{\frac{5}{2}}}{3}\) Is that right?

Answer 9

@ganeshie8

Answer 10

doesn't look correct @StudyGurl14

Answer 11

I used the formula....

Answer 12

\[\dfrac{1}{a-0}\int\limits_0^a a\sqrt{x}\, dx = \dfrac{1}{a} *a*\dfrac{2}{3}a^{3/2} =\dfrac{2}{3}a^{3/2} \]

Answer 13

exponent of \(a\) is just 3/2, not 5/2

Answer 14

Oh, I just noticed I forgot the 1/a-0 part!

Answer 15

Thank you!