Problem: Evaluate the integral

$I = \int_0^1 x^2 e^x \, dx$

and show the step-by-step solution.

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Step-by-Step Solution:

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Step 1: Identify the integration method.
The integral involves the product of a polynomial $x^2$ and an exponential function $e^x$. This suggests that the method of integration by parts is suitable. Recall the formula for integration by parts:

$\int u \, dv = uv - \int v \, du$

We will choose $u = x^2$ (so that differentiation simplifies it) and $dv = e^x dx$ (since the exponential function is easy to integrate).

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Step 2: Compute $u$, $du$, $dv$, and $v$.

• Let $u = x^2$, so $du = 2x \, dx$.
• Let $dv = e^x \, dx$, so $v = e^x$ (since the integral of $e^x$ is itself).

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Step 3: Apply the integration by parts formula.

Using the formula $\int u \, dv = uv - \int v \, du$, we write:

$I = \int_0^1 x^2 e^x \, dx = \left[ x^2 e^x \right]_0^1 - \int_0^1 2x e^x \, dx$

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Step 4: Evaluate the first term $\left[ x^2 e^x \right]_0^1$.

Substitute the limits of integration into $x^2 e^x$:

$\left[ x^2 e^x \right]_0^1 = \left(1^2 e^1\right) - \left(0^2 e^0\right) = e - 0 = e$

Thus, the first term is $e$.

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Step 5: Simplify the remaining integral.

The remaining term is:

$- \int_0^1 2x e^x \, dx$

We now repeat the process of integration by parts for this integral. Let $u = 2x$ and $dv = e^x \, dx$. Then:

• $u = 2x$, so $du = 2 \, dx$
• $dv = e^x \, dx$, so $v = e^x$

Using $\int u \, dv = uv - \int v \, du$, we write:

$\int_0^1 2x e^x \, dx = \left[ 2x e^x \right]_0^1 - \int_0^1 2 e^x \, dx$

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Step 6: Evaluate $\left[ 2x e^x \right]_0^1$.

Substitute the limits into $2x e^x$:

$\left[ 2x e^x \right]_0^1 = \left(2 \cdot 1 \cdot e^1\right) - \left(2 \cdot 0 \cdot e^0\right) = 2e - 0 = 2e$

Thus, the first term is $2e$.

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Step 7: Evaluate $\int_0^1 2 e^x \, dx$.

The integral of $2e^x$ is:

$\int 2e^x \, dx = 2e^x$

Substituting the limits:

$\int_0^1 2 e^x \, dx = \left[ 2e^x \right]_0^1 = \left(2e^1\right) - \left(2e^0\right) = 2e - 2$

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Step 8: Combine all results.

From Step 5, we found:

$\int_0^1 2x e^x \, dx = 2e - (2e - 2) = 2$

Substituting this back into the original integral:

$I = e - \int_0^1 2x e^x \, dx = e - 2$

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Final Answer:

$I = e - 2$

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Do you want further clarification or additional examples?

Related Questions:

1. How does the choice of $u$ and $dv$ affect the integration by parts process?
2. What happens if the limits of integration are changed for this problem?
3. Can we solve this integral using a series expansion for $e^x$?
4. How does integration by parts relate to the product rule in differentiation?
5. Are there alternative methods to solve $\int x^2 e^x \, dx$ without using integration by parts?

Tip:

When applying integration by parts repeatedly, carefully track each substitution to avoid errors in combining terms.