We aim to compute the definite integral:

$I = \int_0^x t \left(e^t - 1\right) \, dt$

Step 1: Expand the integrand

Distribute $t$ across the terms in $e^t - 1$:

$I = \int_0^x t e^t \, dt - \int_0^x t \, dt$

Thus, the integral splits into two parts:

$I = I_1 - I_2$ where: $I_1 = \int_0^x t e^t \, dt, \quad I_2 = \int_0^x t \, dt.$

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Step 2: Solve $I_1 = \int_0^x t e^t \, dt$

This requires integration by parts. Let:

• $u = t$, so $du = dt$,
• $dv = e^t \, dt$, so $v = e^t$.

Using the formula for integration by parts, $\int u \, dv = uv - \int v \, du$, we have:

$I_1 = t e^t \big|_0^x - \int_0^x e^t \, dt.$

Evaluate the second integral:

$\int_0^x e^t \, dt = e^t \big|_0^x = e^x - 1.$

Substitute this result back:

$I_1 = \left[x e^x - 0\right] - \left(e^x - 1\right) = x e^x - e^x + 1.$

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Step 3: Solve $I_2 = \int_0^x t \, dt$

This is a straightforward integral. The antiderivative of $t$ is:

$\int t \, dt = \frac{t^2}{2}.$

Evaluate the definite integral:

$I_2 = \frac{x^2}{2} - \frac{0^2}{2} = \frac{x^2}{2}.$

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Step 4: Combine results

Now, substitute $I_1$ and $I_2$ into $I$:

$I = I_1 - I_2 = \left(x e^x - e^x + 1\right) - \frac{x^2}{2}.$

Simplify:

$I = x e^x - e^x + 1 - \frac{x^2}{2}.$

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Final Answer:

$\int_0^x t \left(e^t - 1\right) \, dt = x e^x - e^x + 1 - \frac{x^2}{2}.$

Would you like further clarification or examples?

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Related Questions:

1. How does integration by parts work in general?
2. Can this integral be solved numerically for specific values of $x$?
3. What is the behavior of $I(x)$ as $x \to \infty$?
4. How can you differentiate $I(x)$ to verify the result?
5. Can you solve this integral using any alternative methods?

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Tip:

Always verify the correctness of a definite integral by differentiating the result with respect to the variable of integration.