Question

given that \[\int\limits_{1}^{3} f(x) dx = -20 and \[\int\limits_{1}^{3} [f(x) +kx^2] dx= 6 find the value of k

Answer 1

You can separate the second into two different integrals:
\[6 =\int\limits\limits_{1}^{3}f(x)dx +\int\limits_{1}^{3}kx^2dx\]
You already know the value of the first integral, so you can substitute the value you have in for that.

Answer 2

\[\int\limits_{1}^{3} (-20+\frac{ k x^{3} }{ 3 }) dx = 6\]

Answer 3

Not quite, you would substitute it like this:
\[6=\int\limits_{1}^{3}f(x)dx + \int\limits_{1}^{3}kx^2dx\]
\[\int\limits_{1}^{3}f(x)dx = -20\]
\[6=(-20)+\int\limits_{1}^{3}kx^2dx\]

Answer 4

ok, after that what? 20+ 6 and become 26 ?

Answer 5

Correct. That then gives you:
\[26=\int\limits_{1}^{3}kx^2dx\]
Remember that constant values can be either kept inside or outside an integral that you are evaluating.

Answer 6

[kx^3/3] is this correct so far?

Answer 7

Yes, that is the indefinite integral of kx^2 (excluding the +c). What does that come out to be when we evaluate it as the definite integral from 1 to 3?

Answer 8

wait, before that, I get this

27k/3 - k/3 = 26.
9k-k/3=26
9k-k=78
8k= 78
k-78/8

this is wrong right? idk how i can get this kind of ending. tell me my mistake

Answer 9

Not quite. So where we are at is:
\[26 = \int\limits_1^3kx^2dx = k\int\limits_1^3x^2dx\]
\[26 = k * \frac{x^3}{3} |_1^3\]
When we evaluate it at those x values, we get something like:
\[26 = k *\left[ \frac{(3)^3}{3} \right]-\left[ \frac{(1)^3}{3} \right]\]

Answer 10

wait, let me try first.

Answer 11

wow, I get the actual answer, k=3 ! tq heril.:D