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
`int3e^(4x)dx`
Answer
`int3e^(4x)dx`
Let `u=4x` then `du=4\ dx`. Our integral becomes:
`int3e^(4x)dx=int3(e^u)(du)/4`
`=3/4inte^udu`
`=3/4e^u+K`
`=3/4e^(4x)+K`
Example 2
`inte^(x^4)4x^3dx`
Answer
`inte^(x^4)4x^3dx`
Let `u=x^4`, then `du=4x^3dx`. Our integral becomes:
`inte^(x^4)4x^3dx=inte^udu`
`=e^u+K`
`=e^(x^4)+K`
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.
Answer
`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)]`
`=3.747`
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`.
Answer
`E=a int_0^(I_0)e^(-Tx)dx`
Let `u = -Tx` then `du = -T\ dx`. Our integral is now:
`E=aint_0^(I_0)e^(-tx)dx`
`=-a/Tint_(x=0)^(x=I_0)e^udu`
`=-a/T[e^u]_(x=0)^(x=I_0)`
`=-a/T[e^(-Tx)]_0^(I_0)`
`=-a/T(e^(-TI_0)-e^0)`
`=a/T(1-e^(-TI_0))`
Exercises
Integrate each of the given functions.
Exercise 1
`int x\ e^(-x^2)dx`
Answer
`intxe^(-x^2)dx`
Put `u = -x^2` then `du = -2x\ dx`.
`int xe^(-x^2)dx=-1/2inte^udu`
`=-1/2e^u+K`
`=-(e^(-x^2))/2+K`
Exercise 2
`int(4\ dx)/(sec x\ e\ ^(sin x)`
Answer
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^-u+K`
`=-4e^(-sin x)+K`
Exercise 3
`int_(-1)^1(dx)/(e^(2-3x))`
Answer
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`.
So
`int_-1^1e^(3x-2)dx=1/3[e^(3x-2)]_-1^1`
`=1/3[e^1-e^-5]`
`=0.9038`
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)`.
Answer
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`
`=2e^(u//2)+K`
`=2e^((x+3)//2)+K`
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:
`y=2e^((x+3)//2)-2e^2`
`=2(sqrt(e^(x+3))-e^2)`
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.)
Answer
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`
`=pi[e^(2x)/2]_0^3`
`=pi[(e^6)/2]-pi[(1)/2]`
`=((e^6-1)/2)pi\ "units"^3`
`=632.1\ "units"^3`
