Question

**Will give medals** Find the value(s) of k such that ..

Answer 1

\[\int\limits_{1}^{2} (2k^2x-3x^2) dx=20\]

Answer 2

lets ignore the constants,

ax + bx^2 is just a polynomial, power rules .... what does it integrate to?

Answer 3

im not sure

Answer 4

if you are doing integration, then you must have done derivatives.

what is a power rule?

Answer 5

how do i take the derivative of x^3 ?

Answer 6

i understand the power rule

Answer 7

like integral x dx = x^2/2 +c

Answer 8

then integrate them ...

Answer 9

ax + bx^2 ... what is our antiderivative?

Answer 10

ax^2/2 +bx^3/3 ?.. i think

Answer 11

correct

now let a = 2k^2 and b=-3

Answer 12

by why are we doing ax+bx^2.....

Answer 13

its easier to process without all the constants cluttering it up

Answer 14

as much as i like that idea, can we just do it like how it is. Its because i have multiple of these and i want to get it right

Answer 15

\[\int\limits_{1}^{2} (\underbrace{(2k^2)}_{a}x+\underbrace{(-3)}_{b} x^2) dx=20\]

we are doing it how it is

\[\int\limits_{1}^{2} (ax+b x^2) dx=20\]
\[\left[\frac a2x^2+\frac b3 x^2\right]_{2}^{1}=20\]

Answer 16

a = 2k^2, b=-3

k^2 x^2 - x^3 evaluted at 1 and 2

Answer 17

where do we go from here?

Answer 18

we have to solve for k so idk, but idk where k is gone too

Answer 19

it hasnt gone anywhere ....

Answer 20

just give me a second while i try to comprehend how this worked

Answer 21

so we should isolate y right?

Answer 22

there is no y

Answer 23

im mean k sorry

Answer 24

hello?

Answer 25

@amistre64

Answer 26

or do we evaluate the integral, the isolate k

Answer 27

3 things at work here, maybe more or less but its a start

integration is linear f(nx+my) = nf(x) + mf(y) for constants n and m

the definite integral of f'(x) is f(b) - f(a), on the interval a to b

algebra is still algebra.

\[\int\limits_{1}^{2} (2k^2x-3 x^2) dx=20\]
\[\int\limits_{1}^{2} (2k^2x)dx+\int\limits_{1}^{2}(-3 x^2) dx=20\]
\[2k^2\int\limits_{1}^{2} (x)dx-3\int\limits_{1}^{2}(x^2) dx=20\]
\[2k^2\left[\frac12x^2\right]_{1}^{2}-3\left[\frac13x^3\right]_{1}^{2}=20\]
\[2k^2\left[\frac12(2^2-1^2)\right]-3\left[\frac13(2^3-1^3)\right]=20\]
\[k^2(2^2-1^2)-(2^3-1^3)=20\]
yes we are solving for k

Answer 28

the answer is 3 correct?

Answer 29

it is actually.

Answer 30

3k^2 - 7 = 20

3k^2 = 27

k^2 has to be 9, so 3 and -3 are fine yes

Answer 31

I have just one more example.

\[\int\limits_{-1}^{0} (5k+3k^2x^2) dx = 6\]

Answer 32

work it, lets see what you do

Answer 33

do i have to split this integral into 2? or is it the same thing again

Answer 34

do whatever is comfortable for you

Answer 35

once we integrate it, it becomes.

\[5k^2/2 + k^2x^3 evaluated at -1, 0 = 6\]

Answer 36

right?

Answer 37

let see if my brain and my fingers can work together ...

5 k x + k^2 x^3

Answer 38

oh your right

Answer 39

lol, eventually i get there

Answer 40

you end up with a quadratic in k

k^2 +5k + 6 = 0

Answer 41

i dont think im doing it right , how did you evaluate it

Answer 42

i end up with 5k-k^2=6

Answer 43

copy the results twice, with a minus between them

5 k x + k^2 x^3
-
5 k x + k^2 x^3


insert the the limits into x


5 k 0 + k^2 0^3
-
5 k -1 + k^2 x^3-1


0
-
-5k-k^2

-(-5k-k^2) = 6)

Answer 44

so i was right?

Answer 45

no, ulness yo have a typo

5k + k^2 = 6

Answer 46

yeah sorry

Answer 47

now how do we isolate y

Answer 48

your working calculus, youve already covered solving a quadratic in algebra classes....

Answer 49

divide by 5 by both sides right?

Answer 50

not really

Answer 51

k^2 +5k -6 = 0

either factor or complete a square, or quadratic formula ... its algebra

Answer 52

nvm im an idiot. thanks man figures it out

Answer 53

good luck :)