Identify Problem: Identify the integral to be solved. We need to evaluate the integral of x ln(x) with respect to x from 0 to 1.Integration by Parts: Use integration by parts. Integration by parts formula is ∫u dv = uv - ∫v du. Let u = ln(x) and dv = x dx. Then we need to find du and v.Find u and v: Differentiate u and integrate dv. Differentiating u with respect to x gives du = (1/x) dx. Integrating dv with respect to x gives v = (1/2)x^2.Differentiate and Integrate: Apply the integration by parts formula. Substitute u, du, and v into the integration by parts formula to get: ∫x ln(x) dx = (ln(x) * (1/2)x^2) - ∫((1/2)x^2 * (1/x) dx).Apply Integration by Parts: Simplify the integral. The integral simplifies to (1/2)x^2 ln(x) - (1/2)∫x dx.Simplify Integral: Evaluate the remaining integral. The integral of x with respect to x is (1/2)x^2, so we have: (1/2)x^2 ln(x) - (1/4)x^2.Evaluate Remaining Integral: Apply the limits of integration from 0 to 1. We need to evaluate (1/2)x^2 ln(x) - (1/4)x^2 from 0 to 1.Apply Limits of Integration: Evaluate the expression at the upper limit. When x = 1, the expression becomes (1/2)(1)^2 ln(1) - (1/4)(1)^2 = 0 - 1/4 = -1/4.Evaluate Upper Limit: Evaluate the expression at the lower limit. When x = 0, we encounter an indeterminate form 0 * ln(0). However, we know that x^2 ln(x) approaches 0 as x approaches 0 from the right. Therefore, the expression becomes 0.Evaluate Lower Limit: Subtract the value at the lower limit from the value at the upper limit. The final value of the integral is -1/4 - 0 = -1/4.

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$\int_{0}^{1}x \ln x\,dx$

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Calculus

Evaluate definite integrals using the chain rule

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\int_{0}^{1}x \ln x\,dx

Identify Problem: Identify the integral to be solved. $\newline$ We need to evaluate the integral of $x \ln(x)$ with respect to $x$ from $0$ to $1$ .
Integration by Parts: Use integration by parts. Integration by parts formula is $\int u \, dv = uv - \int v \, du$ . Let $u = \ln(x)$ and $dv = x \, dx$ . Then we need to find $du$ and $v$ .
Find $u$ and $v$ : Differentiate $u$ and integrate $dv$ . $\newline$ Differentiating $u$ with respect to $x$ gives $du = (1/x) dx$ . $\newline$ Integrating $dv$ with respect to $x$ gives $v = (1/2)x^2$ .
Differentiate and Integrate: Apply the integration by parts formula. $\newline$ Substitute $u$ , $du$ , and $v$ into the integration by parts formula to get: $\newline$ $\int x \ln(x) \, dx = (\ln(x) \cdot (\frac{1}{2})x^2) - \int((\frac{1}{2})x^2 \cdot (\frac{1}{x}) \, dx)$ .
Apply Integration by Parts: Simplify the integral. $\newline$ The integral simplifies to $(\frac{1}{2})x^2 \ln(x) - (\frac{1}{2})\int x \, dx$ .
Simplify Integral: Evaluate the remaining integral. $\newline$ The integral of $x$ with respect to $x$ is $(\frac{1}{2})x^2$ , so we have: $\newline$ $(\frac{1}{2})x^2 \ln(x) - (\frac{1}{4})x^2$ .
Evaluate Remaining Integral: Apply the limits of integration from $0$ to $1$ . We need to evaluate $(\frac{1}{2})x^2 \ln(x) - (\frac{1}{4})x^2$ from $0$ to $1$ .
Apply Limits of Integration: Evaluate the expression at the upper limit. $\newline$ When $x = 1$ , the expression becomes $(\frac{1}{2})(1)^2 \ln(1) - (\frac{1}{4})(1)^2 = 0 - \frac{1}{4} = -\frac{1}{4}$ .
Evaluate Upper Limit: Evaluate the expression at the lower limit. $\newline$ When $x = 0$ , we encounter an indeterminate form $0 \times \ln(0)$ . However, we know that $x^2 \ln(x)$ approaches $0$ as $x$ approaches $0$ from the right. Therefore, the expression becomes $0$ .
Evaluate Lower Limit: Subtract the value at the lower limit from the value at the upper limit. $\newline$ The final value of the integral is $-\frac{1}{4} - 0 = -\frac{1}{4}$ .