Math Problem Statement

could you substitute u with x^2 in that?

Solution

Let's consider the substitution u=x2u = x^2 instead of u=x3u = x^3 for the integral 01x2ex3(1+x3)dx\int_0^1 x^2 e^{x^3} (1 + x^3) \, dx.

First, let's rewrite the integral for clarity:

01x2ex3(1+x3)dx\int_0^1 x^2 e^{x^3} (1 + x^3) \, dx

Now, we'll use the substitution u=x2u = x^2. Then, du=2xdxdu = 2x \, dx, or equivalently, 12du=xdx\frac{1}{2} du = x \, dx.

Next, we need to adjust the limits of integration. When x=0x = 0, u=02=0u = 0^2 = 0. When x=1x = 1, u=12=1u = 1^2 = 1. So the new limits for uu will be from 0 to 1.

Substituting these into the integral, we need to express x2x^2 and x3x^3 in terms of uu. Notice that x2=ux^2 = u and x3=u3/2x^3 = u^{3/2}.

The differential dxdx can be found from 12du=xdx\frac{1}{2} du = x \, dx, so dx=12xdu=12ududx = \frac{1}{2x} du = \frac{1}{2\sqrt{u}} du.

Now, substitute everything into the integral:

01x2ex3(1+x3)dx=01ueu3/2(1+u3/2)12udu\int_0^1 x^2 e^{x^3} (1 + x^3) \, dx = \int_0^1 u e^{u^{3/2}} \left( 1 + u^{3/2} \right) \frac{1}{2\sqrt{u}} \, du

Simplify the integrand:

=1201ueu3/2(1+u3/2)1udu= \frac{1}{2} \int_0^1 u e^{u^{3/2}} \left( 1 + u^{3/2} \right) \frac{1}{\sqrt{u}} \, du

=1201eu3/2(u+u2)du= \frac{1}{2} \int_0^1 e^{u^{3/2}} \left( \sqrt{u} + u^2 \right) \, du

Split the integral into two parts:

=12(01eu3/2udu+01eu3/2u2du)= \frac{1}{2} \left( \int_0^1 e^{u^{3/2}} \sqrt{u} \, du + \int_0^1 e^{u^{3/2}} u^2 \, du \right)

Let’s denote these integrals as I1I_1 and I2I_2:

I1=01eu3/2uduI_1 = \int_0^1 e^{u^{3/2}} \sqrt{u} \, du

I2=01eu3/2u2duI_2 = \int_0^1 e^{u^{3/2}} u^2 \, du

Now, consider the first integral I1I_1:

I1=01eu3/2u1/2duI_1 = \int_0^1 e^{u^{3/2}} u^{1/2} \, du

We use another substitution for this, v=u3/2v = u^{3/2}. Then, dv=32u1/2dudv = \frac{3}{2} u^{1/2} \, du, or du=23u1/2dvdu = \frac{2}{3} u^{-1/2} \, dv.

When u=0u = 0, v=0v = 0. When u=1u = 1, v=13/2=1v = 1^{3/2} = 1.

I1=01ev23u1/2u1/2dv=2301evdvI_1 = \int_0^1 e^v \frac{2}{3} u^{-1/2} u^{1/2} \, dv = \frac{2}{3} \int_0^1 e^v \, dv

=23(ev01)=23(e1)= \frac{2}{3} \left( e^v \bigg|_0^1 \right) = \frac{2}{3} (e - 1)

For the second integral I2I_2:

I2=01eu3/2u2duI_2 = \int_0^1 e^{u^{3/2}} u^2 \, du

Using the same substitution v=u3/2v = u^{3/2}, du=23u1/2dvdu = \frac{2}{3} u^{-1/2} \, dv:

I2=01ev23u4/3u1/2dv=2301evu5/21/2dv=2301evu2dvI_2 = \int_0^1 e^v \frac{2}{3} u^{4/3} u^{-1/2} \, dv = \frac{2}{3} \int_0^1 e^v u^{5/2-1/2} \, dv = \frac{2}{3} \int_0^1 e^v u^2 \, dv

Since u=v2/3u = v^{2/3}:

=2301evv4/3dv= \frac{2}{3} \int_0^1 e^v v^{4/3} \, dv

This integral does not simplify easily, so we stick with the original substitution u=x3u = x^3 for a straightforward solution.

Thus, the exact value of the integral remains:

13e\boxed{\frac{1}{3} e}

Would you like further details or have any questions?

Here are 5 more questions you might consider:

  1. How do you perform integration by parts?
  2. Can you explain substitution in integrals in more detail?
  3. How do you solve definite integrals with variable limits?
  4. What are some common techniques for evaluating complex integrals?
  5. Can you solve xsin(x2)dx\int x \sin(x^2) \, dx?

Tip: When performing substitutions, carefully track how variables and differentials change to ensure accurate results.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Substitution in Integrals
Definite Integrals

Formulas

Integral of x^2 e^(x^3) (1 + x^3) dx
Substitution u = x^2

Theorems

Substitution Method for Integrals

Suitable Grade Level

Grades 11-12