Expert: Paul Klarreich - 11/10/2006

Question
Paul I got just a few left on my study guide and I know you will probably find these easy but I don't yet.  Thanks for your help so far.  You have been a blessing.

Find the positive value of k such that the area under the graph of y=e^3x over the interval [0,k] is 3 sq units.

Thanks in advance.

Answer
Questioner:  Gary
Category:  Calculus
 
Question:  Paul I got just a few left on my study guide and I know you will probably find these easy but I don't yet.  Thanks for your help so far.  You have been a blessing.

Find the positive value of k such that the area under the graph of y=e^3x over the interval [0,k] is 3 sq units.

Thanks in advance.
......................................
Hi, Gary,

Your area should be:

(k
|   e^3x dx
)0

Now there is a basic rule about these things that are called DEFINITE INTEGRALS.  The variable of integration, here x, is a DUMMY VARIABLE.  That's a way of saying that the expression above is NOT A FUNCTION OF X.  It has nothing to do with x at all.  In fact, it is a function of k.  If you had something like:

(5
|   e^3x dx
)0

Then what you have is just a number.  The symbol 'x' will disappear by the end, having been replaced by the 0 and the 5.

Working your problem:

(k
|   e^3x dx
)0

= e^3x/3  from  0 to k

= e^3k/3 - e^0

= e^3k/3 - 1

Now you want to set that equal to your 3 sq units:

e^3k/3 - 1 = 3

e^3k/3  = 4

e^3k  = 12

3k = ln 12
k = (ln 12)/3