Find a positive value of k such that...?

Question

....such that the area of the graph under y=e^(5x) over the interval [0,k] is 6 square units.

Thanks :)

Answer 1

We want to find k so that ∫(e^(5x)) dx on [0, k] = 6. Let's integrate and then figure out what k has to be.

An antiderivative of e^(5x) is (1/5)e^(5x). Evaluating from 0 to k gives (1/5)e^(5k) - (1/5)e^0 = (1/5)e^(5k) - 1.

We want (1/5)e^(5k) - 1 = 6. Solve:

(1/5)e^(5k) = 7
e^(5k) = 35
5k = ln (35)
k = (ln 35)/5.

Answer 2

The area of the graph y = e^5x over the interval [0,k] is the integral of y from 0 to k. Since it is equal to 6 sq. units, then,
∫(0,k) e^5x dx = 6

The integral of e^5x = 1/5 e^5x. Therefore,
(1/5 e^5x)|(0,k) =6

Thus,
1/5 e^(5k) - 1/5 e^(5·0) = 6

Then, we can now solve for k.
Multiply the equation by 5
e^5k - e^0 = 30

We simplify e^0 = 1
e^5k - 1 = 30

We transpose:
e^5k = 31

Then, we take the ln (natural logarithm) of both sides.
ln e^5k = ln 31

Thus, using the property of logarithms log xª = a log x
5k ln e= ln 31

Since ln e = 1, then
5k = ln 31

Therefore,
k = 1/5 ln 31 ≈ 0.68679744...

^_^

Answer 3

Call this integral f(k), so

f(k) = Integral (e^(5x) dx) , 0 to k

Now the integral of e^x is e^x, so the integral of e^(5x) is e^(5x)/5:

f(k) = [e^(5x) / 5], 0 to k
= (e^(5k) - 1) / 5.

We want f(k) = 6, so
(e^(5k) - 1) / 5 = 6
so e^(5k) = (6*5)+1 = 31
so 5k = ln 31
and thus k = (ln 31) / 5.

Answer 4

6 = 0.2 e^(5k) - 0.2
30 = e^(5k) - 1
31 = e^(5k)
ln(31) = 5k
k = 0.2 ln(31) = 0.6868

Answer 5

KECK !