Integral of e

[Jason76]

Bad thought process? Most texts write this differently.

Strategy

Find integral of e by using formula: \(\displaystyle \int e^{x} dx\) or \(\displaystyle \int ae^{x} dx \) = \(\displaystyle e^{x} * dx\) by using information like a, b, u, du and dx. Problem starts in either as \(\displaystyle \int e^{x} dx\) or as \(\displaystyle \int ae^{x} dx\) (constant a included)

1. Find du from u. Use substitution: The problem becomes \(\displaystyle e^{u} dx\) or \(\displaystyle ae^{u} dx\)

2. Solve for dx when problem is in form of du = bdx. b = number that's not a constant.

3. Multiply dx and the constant (if it exists). Product is still dx.

4. Use final formula.

Example 1

What is \(\displaystyle \int e^{7x - 3} dx\) ?

\(\displaystyle u = 7x - 3\)

\(\displaystyle du = 7\)

or

\(\displaystyle du = 7dx\)

Solving for dx

\(\displaystyle dx = \dfrac{1}{7}du\) (Next, make du disappear without math.)

Answer would be:

\(\displaystyle \dfrac{1}{7} e^{7x - 3} + C\) (Note: This example had no constant. What if a constant a is added?)

Example 2

What is \(\displaystyle \int5e^{7x - 3}dx\) ?

\(\displaystyle u = 7x - 3\)

\(\displaystyle du = 7\)

or

\(\displaystyle du = 7dx\)

Solving for dx

\(\displaystyle dx = \dfrac{1}{7}du\) (Next, make du symbol disappear without math.)

Extra step (Multiplying dx by constant in original problem - product is still considered dx)

\(\displaystyle \dfrac{1}{7} * 5 * e^{8x - 3}\)

Final Answer:

\(\displaystyle \dfrac{5}{7} e^{7x - 3} + C\)

Checking

Now you can check by taking the derivative of the answer (ignoring C by "plugging in 0 for C"). The answer would be the original integration problem.

Other Applications

If trying to find other integrals of the form \(\displaystyle \int x dx\) or \(\displaystyle \int ax dx\) then a similar strategy is used.

 

[HallsofIvy]

Bad thought process? Most texts write this differently.

Strategy

Find integral of e by using formula: \(\displaystyle \int e^{x} dx\) or \(\displaystyle \int ae^{x} dx \) = \(\displaystyle e^{x} * dx\)

There should not be a "dx" here- that is part of the integration symbol. Perhaps what you meant was \(\displaystyle = e^x+ C\) where C is any constant.

by using information like a, b, u, du and dx. Problem starts in either as \(\displaystyle \int e^{x} dx\) or as \(\displaystyle \int ae^{x} dx\) (constant a included)

1. Find du from u. Use substitution: The problem becomes \(\displaystyle e^{u} dx\) or \(\displaystyle ae^{u} dx\)

No, after converting to variable "u" your integral must have "du", not "dx".

2. Solve for dx when problem is in form of du = bdx. b = number that's not a constant.

This makes no sense at all until you tell us what "u" is! If, for example, the problem were \(\displaystyle \int e^{a+ bx}dx\), then you can simplify it by making the substitution u= a+ bx. Then, differentiating, du/dx= b so du= bdx and dx= (1/b)du. With that substitution, the integral becomes \(\displaystyle \frac{1}{b}\int e^u du\). "b= number that's not a constant" also makes no sense. A "number" is a "constant" pretty much by definition.

3. Multiply dx and the constant (if it exists). Product is still dx.

4. Use final formula.

Example 1

What is \(\displaystyle \int e^{7x - 3} dx\) ?

\(\displaystyle u = 7x - 3\)

\(\displaystyle du = 7\)

or

\(\displaystyle du = 7dx\)

Yes, this is correct, not "du= 7"

Solving for dx

\(\displaystyle dx = \dfrac{1}{7}du\) (Next, make du disappear without math.)

I have no idea what "make du disappear without math" means.
If you mean "integrate", yes, \(\displaystyle \frac{1}{7}\int e^u du= \frac{1}{7}e^u+ C\)

Answer would be:

\(\displaystyle \dfrac{1}{7} e^{7x - 3} + C\) (Note: This example had no constant. What if a constant a is added?)

Okay, replacing u by 7x- 3 again gives that. But what do you mean "had no constant"? That "C" is the "constant of integration".

Example 2

What is \(\displaystyle \int5e^{7x - 3}dx\) ?

\(\displaystyle u = 7x - 3\)

\(\displaystyle du = 7\)

or

\(\displaystyle du = 7dx\)

Again, "du= 7" is not correct. You cannot have a differential equal to a number or, in general, any function of x. There must be a "dx" on the right side.

Solving for dx

\(\displaystyle dx = \dfrac{1}{7}du\) (Next, make du symbol disappear without math.)

Extra step (Multiplying dx by constant in original problem - product is still considered dx)

\(\displaystyle \dfrac{1}{7} * 5 * e^{8x - 3}\)

Final Answer:

\(\displaystyle \dfrac{5}{7} e^{7x - 3} + C\)

Yes. You learned, well before you learned about "substitution", that \(\displaystyle \int c f(x)dx= c\int f(x)dx\). So \(\displaystyle \int 5e^{7x- 3}dx= 5\int e^{7x- 3}dx\). Since you have already determined that \(\displaystyle \int e^{7x- 3}dx= \frac{1}{7}e^{7x- 3}+ C\), you can immediately write \(\displaystyle \int 5e^{7x- 3}dx= 5(\frac{1}{7}e^{7x- 3}+ C= \frac{5}{7}e^{7x- 3}+ 5C\). Since "C" can be any number, so is "5C" and you can just write "C" to mean that new constant.

Checking

Now you can check by taking the derivative of the answer (ignoring C by "plugging in 0 for C"). The answer would be the original integration problem.

There is no reason to "plug in 0 for C", the derivative of any constant is 0. Yes, the derivative of \(\displaystyle \frac{5}{7}e^{7x- 3}+ C\) is \(\displaystyle \frac{5}{7}(7)e^{7x- 3}+ 0= 5e^{7x- 3}\), the function you started with.

Other Applications

If trying to find other integrals of the form \(\displaystyle \int x dx\) or \(\displaystyle \int ax dx\) then a similar strategy is used.

I'm not sure what you mean here. "\(\displaystyle \int x dx\)" and \(\displaystyle \int ax dx\) don't need a "strategy"- they are trivial. If you mean \(\displaystyle \int f(x)dx\) and \(\displaystyle \int af(x)dx\) for any functions f(x), that's much too general. You typically need to learn a wide variety of "strategies" to integrate a variety of functions. And, there exist perfectly reasonable functions, such as \(\displaystyle e^{x^2}\), which have integrals (anti-derivatives) but those anti-derivatives cannot be written in terms of "elementary functions".