The integral you provided is:
β«x3e2xdx
To solve this, we use integration by parts multiple times. Integration by parts is based on the formula:
β«udv=uvββ«vdu
Step 1: Choose u and dv
Let:
- u=x3, so du=3x2dx
- dv=e2xdx, so v=21βe2x (since the derivative of e2x is 2e2x)
Step 2: Apply Integration by Parts
Now apply the integration by parts formula:
β«x3e2xdx=21βx3e2xββ«21β3x2e2xdx
Simplify:
=21βx3e2xβ23ββ«x2e2xdx
Step 3: Repeat the process for β«x2e2xdx
Let:
- u=x2, so du=2xdx
- dv=e2xdx, so v=21βe2x
Now apply integration by parts again:
β«x2e2xdx=21βx2e2xββ«21β2xe2xdx
Simplify:
=21βx2e2xβ21ββ«2xe2xdx
Step 4: Repeat for β«xe2xdx
Let:
- u=x, so du=dx
- dv=e2xdx, so v=21βe2x
Now apply integration by parts:
β«xe2xdx=21βxe2xβ21ββ«e2xdx
The last integral is straightforward:
β«e2xdx=21βe2x
So:
β«xe2xdx=21βxe2xβ41βe2x
Step 5: Combine all results
Now substitute back into the original equation:
β«x2e2xdx=21βx2e2xβ21β(21βxe2xβ41βe2x)
Simplify:
=21βx2e2xβ41βxe2x+81βe2x
Finally, substitute this back into the previous result:
β«x3e2xdx=21βx3e2xβ23β(21βx2e2xβ41βxe2x+81βe2x)
Simplify:
β«x3e2xdx=21βx3e2xβ43βx2e2x+83βxe2xβ163βe2x+C
Final Answer:
β«x3e2xdx=21βx3e2xβ43βx2e2x+83βxe2xβ163βe2x+C
Would you like any further details or clarification on any step?
Here are 5 related questions:
- How do you choose the u and dv in integration by parts?
- Can integration by parts be used for other transcendental functions like sin(x)?
- What is the general formula for integration by parts?
- How does repeated integration by parts work in practical applications?
- What are the conditions under which integration by parts simplifies a problem?
Tip: When integrating by parts multiple times, always check for a recursive pattern to simplify calculations.