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3. Integration: The Exponential Form

by M. Bourne

By reversing the process in obtaining the derivative of the exponential function, we obtain the remarkable result:

`int e^udu=e^u+K`

It is remarkable because the integral is the same as the expression we started with. That is, `e^u`.

Example 1




Let `u=4x` then `du=4\ dx`. Our integral becomes:





Example 2




Let `u=x^4`, then `du=4x^3dx`. Our integral becomes:




Example 3

`int_0^1 sec^2x e^(tan x)dx`

Here's the curve `y=sec^2x e^(tan x)`:

The shaded region represents the integral we need to find.


`int_0^1sec^2 x\ e^(tan x)dx`

Let `u=tan x`, then `du=sec^2 x\ dx`. So we have:

`int_0^1sec^2x\ e^(tan x)dx=[e^(tan x)]_0^1`

`=[e^(tan 1)]-[e^(tan 0)]`


Of course, `x` is in radians. These integration techniques don't work in degrees.

Example 4

In the theory of lasers, we see

`E=a int_0^(I_0)e^(-Tx)dx`

where `a`, `I_0` and `T` are constants. Find `E`.


`E=a int_0^(I_0)e^(-Tx)dx`

Let `u = -Tx` then `du = -T\ dx`. Our integral is now:








Integrate each of the given functions.

Exercise 1

`int x\ e^(-x^2)dx`



Put `u = -x^2` then `du = -2x\ dx`.

`int xe^(-x^2)dx=-1/2inte^udu`



Exercise 2

`int(4\ dx)/(sec x\ e\ ^(sin x)`


Since `1/(sec x)=cos x`, we can re-write the question as:

`int(4\ cos x\ dx)/(e^(sin x)`

Put `u = sin x` then `du = cos x\ dx`

`int(4\ cos x\ dx)/(e^(sin x))=4int(du)/e^u`

`=4int e^-u\ du`


`=-4e^(-sin x)+K`

Exercise 3



Since `−(2 − 3x) = 3x − 2`, we can bring the denominator to the top and write the question as:

`int_(-1)^1 e^(-(2-3x)) dx =int_-1^1e^(3x-2)dx`

Put `u = 3x − 2` then `du = 3\ dx`.





Here is the curve `y=1/e^(2-3x)`:

The shaded region represents the integral we just found.

Exercise 4

Find the equation of the curve for which `(dy)/(dx)=sqrt(e^(x+3))` if the curve passes through `(1, 0)`.


We need to find

`y=int sqrt(e^(x+3))dx,`

and subsitute our given conditions to find the equation of the curve.

Put `u = x + 3` then `du = dx`. Perform the integral.

`y=intsqrt(e^(x+3)) dx`

`=intsqrt(e^u) du`

`=inte^(u//2) du`



Now, the curve passes through `(1, 0)`.

This means when `x = 1`, `y = 0`.

So `0=2e^2+K`, giving `K = -2e^2`.

So the required equation of the curve is:



The graph of the solution curve we just found, showing that it passes through (1, 0).

Application - Volume of Solid of Revolution

The area bounded by the curve `y = e^x`, the `x`-axis and the limits of `x = 0` and `x = 3` is rotated about the `x`-axis. Find the volume of the solid formed. (You may wish to remind yourself of the volume of solid of revolution formula.)


The graph of `y=e^x`, with the area under the curve between `x=0` to `x=3` shaded.

When the shaded area is rotated 360° about the x-axis, we have:

Area under the curve `y=e^x` from `x=0` to `x=3` rotated around the `x`-axis.

Applying the volume of a solid of revolution formula, we get

`V=pi int_a^by^2dx`

`=pi int_0^3(e^x)^2dx`

`=pi int_0^3e^(2x)dx`



`=((e^6-1)/2)pi\ "units"^3`

`=632.1\ "units"^3`

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