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Chapter 7: Problem 34

Evaluate the following definite integrals. $$\int_{0}^{\ln 2} x e^{x} d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the product between x and e^x from 0 to ln(2). Answer: The definite integral is equal to 2(ln(2) - 1).

Step by step solution

01

Apply Integration by Parts

Integration by Parts states that for two functions u and v: $$\int u \cdot d v = u \cdot v - \int v \cdot d u$$ Choose: $$u = x, d v = e^x d x$$ Differentiate u and integrate d v to find d u and v: $$d u = d x, v = e^x$$
02

Use the Formula

Now substitute u, d u, v, and d v into the Integration by Parts formula: $$\int_{0}^{\ln 2} x e^{x} d x = \int u \cdot d v = u\cdot v - \int v \cdot d u$$ Insert the expressions for u, v, d u, and d v into the formula: $$\int_{0}^{\ln 2} x e^{x} d x = x e^x - \int e^x d x$$
03

Integrate and Evaluate

Now we integrate and evaluate definite integral: $$\int_{0}^{\ln 2} x e^{x} d x = \left[ x e^x - e^x\right]_0^{\ln 2}$$ Evaluate at the upper and lower limit: $$\left[ (\ln 2)e^{\ln 2} - e^{\ln 2}\right] - \left[0\right] = (\ln 2 - 1)e^{\ln 2}$$ Since \(e^{\ln 2} = 2\), the final answer is: $$\int_{0}^{\ln 2} x e^{x} d x = (\ln 2 - 1)2 = 2(\ln 2 - 1)$$

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