Question

calculus question

Answer 1

how do i start?

Answer 2

derive every term

Answer 3

\[\large 4+\int _A^x\frac{g(t)}{t^2}dt=2\sqrt{2}\]
for the integral use FTC 2

Answer 4

i triedusing the fundamental theorem of calc, but it didnt work

Answer 5

\[\int_a^bf(t)dt=F(b)-F(a)\]

Answer 6

how can i integrate \[ \frac{g(t)}{t^2}\]?

Answer 7

dont intergrate
for example \[\large\text{the derivative of }\int _0^xf(t)dt=f(x)-f(0)\]

Answer 8

but dont you have to integrate the function first, for FTC 2?

Answer 9

sry i meant ftc 1

Answer 10

if its one, then it would \[ \frac{g(x)}{x^2}\]
but still wont work..

Answer 11

it doesnt matter how complex the function to be integrated is,all we gotta do is jus write that function with A and x substituted....inside
\[\large \int _A^x\frac{g(t)}{t^2}dt=\frac{g(x)}{x}-\frac{g(A)}{A^2}\]
so since the derivative of 4 and \[2\sqrt{2}\] is 0 we have
\[\frac{g(x)}{x}-\frac{g(A)}{A^2}=0\]
\[g(x)=\frac{xg(A)}{A^2}\]

Answer 12

ohh wait the other side is not \[2\sqrt{2}\]its \[2\sqrt{x}\]
so derivative is
\[\frac{1}{\sqrt{x}}\]

Answer 13

so we have
\[\frac{g(x)}{x^2}-\frac{g(A)}{A^2}=\frac{1}{\sqrt{x}}\implies g(x)=\frac{x^2}{\sqrt{x}}+\frac{x^2g(A)}{A^2}\]

Answer 14

how am i able to find the function of g and A?

Answer 15

the derivative of \[\int_A^x \frac{g(t)}{t^2}dt\] is \[\frac{g(x)}{x^2}\], not \[\frac{g(x)}{x^2} - \frac{g(A)}{A^2}\]

Answer 16

so \[ g(x) = x^{3/2}\]?

Answer 17

that's right

Answer 18

how am i able to find \(A\)?

Answer 19

wait does it mean \[g(t) = t^{3/2}\]?

Answer 20

now, you have to evaluate the definite integral

Answer 21

\[4+\int_A^x\frac{t^{3/2}}{t^2}dt=2\sqrt x\]

Answer 22

i get \(A = 4\) is that right?

Answer 23

yup, that is correct

Answer 24

yay! thank you! :D

Answer 25

thank you too.