Question

determine the srea of the region bounded by f(x)=8x^2+4 and the x-axis is between 2 and 3

Answer 1

just integrate f(x) from 2 to 3

Answer 2

can you show me step by step, im really confused

Answer 3

as far as im reading it right .... do you know what integration is?

Answer 4

is it the anti derivative

Answer 5

correct

so, whatis the antiD of 8x^2? and what it is it of 4?

Answer 6

(8/3)x^3?

Answer 7

good, and the one for the 4?

Answer 8

4x

Answer 9

the usual notation for integration is;
\[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]

you just found the F(x) parts soo

\[\int_{2}^{3}8x^2+4~dx=(\frac83(3)^3+4(3))-(\frac83(2)^3+4(2))\]

Answer 10

so the answer is 164/3

Answer 11

thats what i get

Answer 12

now one more question

Answer 13

determine the region bounded by f(x)=\[4/\sqrt[2]{x}\] and the x axis between 16 and 25

Answer 14

turn it into a power and follow the same rules you did for this one

Answer 15

the anti derivitve of 4 will be 4x^2

Answer 16

\[\frac{4}{\sqrt x}\]
\[\frac{4}{x^{1/2}}\]
\[f(x)=4x^{-1/2}\]

Answer 17

ohh quotient rule?

Answer 18

no, its just the reverse of the power rule, just like you did for that 8x^2

Answer 19

add 1 and divide

Answer 20

ooo

Answer 21

\[\int kx^n=\frac {kx^{n+1}}{n+1} \]

or recall\[\frac d{dx}(kx^n)=kn~x^{n-1}\]
\[\int kn~x^{n-1}=\frac {kn~x^{n-1+1}}{n-1+1} =\frac {k\cancel{n}~x^{n}}{\cancel{n}}=kx^n\]

Answer 22

answer =8

Answer 23

8sqrt(25)-8sqrt(16)

8(5-4) ; i agree, its 8 :)