Integral of (1-7x)e^(5x) dx
The solution
Detail solution
-
There are multiple ways to do this integral.
Method #1
-
Rewrite the integrand:
(1−7x)e5x=−7xe5x+e5x
-
Integrate term-by-term:
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−7xe5x)dx=−7∫xe5xdx
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=x and let dv(x)=e5x.
Then du(x)=1.
To find v(x):
-
There are multiple ways to do this integral.
Method #1
-
Let u=5x.
Then let du=5dx and substitute 5du:
∫25eudu
-
The integral of a constant times a function is the constant times the integral of the function:
∫5eudu=5∫eudu
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 5eu
Now substitute u back in:
5e5x
Method #2
-
Let u=e5x.
Then let du=5e5xdx and substitute 5du:
∫251du
-
The integral of a constant times a function is the constant times the integral of the function:
∫51du=5∫1du
-
The integral of a constant is the constant times the variable of integration:
∫1du=u
So, the result is: 5u
Now substitute u back in:
5e5x
Now evaluate the sub-integral.
-
The integral of a constant times a function is the constant times the integral of the function:
∫5e5xdx=5∫e5xdx
-
Let u=5x.
Then let du=5dx and substitute 5du:
∫25eudu
-
The integral of a constant times a function is the constant times the integral of the function:
∫5eudu=5∫eudu
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 5eu
Now substitute u back in:
5e5x
So, the result is: 25e5x
So, the result is: −57xe5x+257e5x
-
Let u=5x.
Then let du=5dx and substitute 5du:
∫25eudu
-
The integral of a constant times a function is the constant times the integral of the function:
∫5eudu=5∫eudu
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 5eu
Now substitute u back in:
5e5x
The result is: −57xe5x+2512e5x
Method #2
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=1−7x and let dv(x)=e5x.
Then du(x)=−7.
To find v(x):
-
Let u=5x.
Then let du=5dx and substitute 5du:
∫25eudu
-
The integral of a constant times a function is the constant times the integral of the function:
∫5eudu=5∫eudu
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 5eu
Now substitute u back in:
5e5x
Now evaluate the sub-integral.
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−57e5x)dx=−57∫e5xdx
-
Let u=5x.
Then let du=5dx and substitute 5du:
∫25eudu
-
The integral of a constant times a function is the constant times the integral of the function:
∫5eudu=5∫eudu
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 5eu
Now substitute u back in:
5e5x
So, the result is: −257e5x
Method #3
-
Rewrite the integrand:
(1−7x)e5x=−7xe5x+e5x
-
Integrate term-by-term:
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−7xe5x)dx=−7∫xe5xdx
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=x and let dv(x)=e5x.
Then du(x)=1.
To find v(x):
-
Let u=5x.
Then let du=5dx and substitute 5du:
∫25eudu
-
The integral of a constant times a function is the constant times the integral of the function:
∫5eudu=5∫eudu
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 5eu
Now substitute u back in:
5e5x
Now evaluate the sub-integral.
-
The integral of a constant times a function is the constant times the integral of the function:
∫5e5xdx=5∫e5xdx
-
Let u=5x.
Then let du=5dx and substitute 5du:
∫25eudu
-
The integral of a constant times a function is the constant times the integral of the function:
∫5eudu=5∫eudu
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 5eu
Now substitute u back in:
5e5x
So, the result is: 25e5x
So, the result is: −57xe5x+257e5x
-
Let u=5x.
Then let du=5dx and substitute 5du:
∫25eudu
-
The integral of a constant times a function is the constant times the integral of the function:
∫5eudu=5∫eudu
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 5eu
Now substitute u back in:
5e5x
The result is: −57xe5x+2512e5x
-
Now simplify:
25(12−35x)e5x
-
Add the constant of integration:
25(12−35x)e5x+constant
The answer is:
25(12−35x)e5x+constant
The answer (Indefinite)
[src]
/
| 5*x 5*x
| 5*x 12*e 7*x*e
| (1 - 7*x)*e dx = C + ------- - --------
| 25 5
/
∫(1−7x)e5xdx=C−57xe5x+2512e5x
The graph
5
12 23*e
- -- - -----
25 25
−2523e5−2512
=
5
12 23*e
- -- - -----
25 25
−2523e5−2512
Use the examples entering the upper and lower limits of integration.