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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

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

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.

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.

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

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)

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))

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).

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.)

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