Let's consider the substitution u=x2 instead of u=x3 for the integral ∫01x2ex3(1+x3)dx.
First, let's rewrite the integral for clarity:
∫01x2ex3(1+x3)dx
Now, we'll use the substitution u=x2. Then, du=2xdx, or equivalently, 21du=xdx.
Next, we need to adjust the limits of integration. When x=0, u=02=0. When x=1, u=12=1. So the new limits for u will be from 0 to 1.
Substituting these into the integral, we need to express x2 and x3 in terms of u. Notice that x2=u and x3=u3/2.
The differential dx can be found from 21du=xdx, so dx=2x1du=2u1du.
Now, substitute everything into the integral:
∫01x2ex3(1+x3)dx=∫01ueu3/2(1+u3/2)2u1du
Simplify the integrand:
=21∫01ueu3/2(1+u3/2)u1du
=21∫01eu3/2(u+u2)du
Split the integral into two parts:
=21(∫01eu3/2udu+∫01eu3/2u2du)
Let’s denote these integrals as I1 and I2:
I1=∫01eu3/2udu
I2=∫01eu3/2u2du
Now, consider the first integral I1:
I1=∫01eu3/2u1/2du
We use another substitution for this, v=u3/2. Then, dv=23u1/2du, or du=32u−1/2dv.
When u=0, v=0. When u=1, v=13/2=1.
I1=∫01ev32u−1/2u1/2dv=32∫01evdv
=32(ev01)=32(e−1)
For the second integral I2:
I2=∫01eu3/2u2du
Using the same substitution v=u3/2, du=32u−1/2dv:
I2=∫01ev32u4/3u−1/2dv=32∫01evu5/2−1/2dv=32∫01evu2dv
Since u=v2/3:
=32∫01evv4/3dv
This integral does not simplify easily, so we stick with the original substitution u=x3 for a straightforward solution.
Thus, the exact value of the integral remains:
31e
Would you like further details or have any questions?
Here are 5 more questions you might consider:
- How do you perform integration by parts?
- Can you explain substitution in integrals in more detail?
- How do you solve definite integrals with variable limits?
- What are some common techniques for evaluating complex integrals?
- Can you solve ∫xsin(x2)dx?
Tip: When performing substitutions, carefully track how variables and differentials change to ensure accurate results.