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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

`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.

Continues below

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))`

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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`

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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`

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