Question

find a function f(x) and a number a such that 4 + integral from a to x^(2) f(t)/t dt = 2sqt x

Answer 1

\[4+\int\limits_{a}^{x^2}\frac{f(t)}{t} dt =2 \sqrt{x} ?\]

Answer 2

let's try something
\[\text{ Let } g(t) =\frac{f(t)}{t} \text{ let G(t) be antiderivative} \]
\[4 +\int\limits_{a}^{x^2}g(t) dt =2 \sqrt{x}\]
\[4+G(t)|_a^{x^2} =2 \sqrt{x}\]
\[4+G(x^2)+G(a)=2 \sqrt{x}\]
now if we differentiate both sides we get
\[0+(x^2)' g(x^2)+0=\frac{1}{\sqrt{x}}\]
\[2xg(x^2)=\frac{1}{\sqrt{x}} => g(x^2)=\frac{1}{ \sqrt{x}} \cdot \frac{1}{2x}\]
\[g(x^2)=\frac{1}{2x \sqrt{x}}\]
\[\frac{f(x^2)}{x^2}=\frac{1}{2 x \sqrt{x}}\]
\[f(x^2)=\frac{x^2}{2 x \sqrt{x}}\]
\[f(x^2)=\frac{x^2}{2 x^{1+\frac{1}{2}}}=\frac{x^2}{2 x^{\frac{3}{2}}}\]
\[f(x^2)=\frac{(x)^2}{2 ((x)^2)^\frac{1}{2} \cdot ((x)^2)^\frac{1}{4}}\]

Answer 3

\[f(x)=\frac{x}{2 x ^\frac{1}{2} \cdot x ^\frac{1}{4}}\]

Answer 4

so \[f(x)=\frac{x}{2 x^\frac{3}{4}}=\frac{1}{2}x^{1-\frac{3}{4}}=\frac{1}{2}x^{\frac{1}{4}}\]

Answer 5

\[4+\int\limits_{a}^{x^2} \frac{1}{2} \cdot \frac{t^\frac{1}{4}}{t} dt =2 \sqrt{x}\]
\[4+\frac{1}{2} \int\limits_{a}^{x^2}t^{\frac{-3}{4}} dt = 2 \sqrt{x}\]
\[4+\frac{1}{2} \cdot \frac{t^{\frac{-3}{4}+1}}{\frac{-3}{4}+1}|_a^{x^2}=2 \sqrt{x}\]
\[4+\frac{1}{2} \cdot \frac{t^\frac{1}{4}}{\frac{1}{4}}|_a^{x^2}=2 \sqrt{x}\]
\[4+\frac{1}{2} \cdot 4[(x^2)^\frac{1}{4}-a^\frac{1}{4}]=2 \sqrt{x}\]
\[4+2[x^\frac{1}{2}-a^\frac{1}{4}] =2 \sqrt{x}\]
\[2 \sqrt{x} +4 -a^\frac{1}{4}= 2 \sqrt{x}\]

Answer 6

\[=> 4- a^\frac{1}{4}=0\]

Answer 7

solve for a

Answer 8

\[4 =a^\frac{1}{4}\]

Answer 9

\[4^4=a\]

Answer 10

this is a cute problem

Answer 11

i should give this to my cal students

Answer 12

you have a small typo...

Answer 13

no!

Answer 14

after

\[4+2[x^\frac{1}{2}-a^\frac{1}{4}] =2 \sqrt{x}\]

Answer 15

darn it stupid 2

Answer 16

\[4-2 a^\frac{1}{4}=0\]

Answer 17

\[4=2 a^\frac{1}{4}\]

Answer 18

\[2=a^\frac{1}{4}\]

Answer 19

\[2^4=a\]

Answer 20

nice problem

Answer 21

did i go the long way?

Answer 22

not that I can see

Answer 23

ok it does seem to be lengthy
it should be bonus
they probably won't figure out anyways

Answer 24

prob not...probably get stuck at the f(x^2) part


... nice work

Answer 25

yes thats what i was thinking