Question

What are the values of k for which:

Answer 1

\[\int\limits_{-3}^{k}x^2dx=0\]

Answer 2

one value should be more or less obvious
do you know it ?

Answer 3

3

Answer 4

No, actually. not three

Answer 5

hmm no i don't think 3 works
this function is always postie so any integral will also be positive unlesss

Answer 6

"positive"

Answer 7

k = 3 would work only if the function was odd

but instead, you use this rule

\[\Large \int_{k}^{k}f(x)dx = 0\]

Answer 8

It'd have to be -3

Answer 9

the integral I wrote represents the area under the curve from x = k to x = k, which is an infinitely thin rectangle with 0 area

Answer 10

yes, k = -3

Answer 11

And that would have to be the only value.

Answer 12

yes because x^2 is always positive (well the nonzero x values anyway) and never dips below the x axis

Answer 13

Makes sense.

Answer 14

Here is another way to solve


Let


\[\Large g(x) = \int x^2dx\]


You'll find, after integrating, that


\[\Large g(x) = \frac{x^3}{3}+C\] where C is any constant


-------------------------


\[\Large g(x) = \frac{x^3}{3}+C\]
\[\Large g(k) = \frac{k^3}{3}+C\]


-------------------------


\[\Large g(x) = \frac{x^3}{3}+C\]
\[\Large g(-3) = \frac{(-3)^3}{3}+C\]
\[\Large g(-3) = \frac{-27}{3}+C\]

-------------------------


so,


\[\Large \int_{-3}^{k}x^2dx = g(k) - g(-3)\]


\[\Large \int_{-3}^{k}x^2dx = \left(\frac{k^3}{3}+C\right)- \left(\frac{-27}{3}+C\right) = 0\]

I'm going to skip a bunch of steps, but solving `(1/3)k^3+C - (-27/3)-C = 0` for k leads to k = -3. Technically there are 2 other solutions, but they are not real numbers. So we can ignore them.