Skip to main content
Search IntMath
Close

450+ Math Lessons written by Math Professors and Teachers

5 Million+ Students Helped Each Year

1200+ Articles Written by Math Educators and Enthusiasts

Simplifying and Teaching Math for Over 23 Years

Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.

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`

Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.