Question

Use integration by parts to evaluate the following integrals. $$\int_{0}^{1} x \ln x d x$$

Short answer

Short Answer: The definite integral of the given function $$\int_{0}^{1} x \ln x d x$$ is equal to $$-\frac{1}{4}$$.

Step-by-step solution

Identify u and dv

In this exercise, the integral we need to evaluate is: $$\int_{0}^{1} x \ln x d x$$ We will use integration by parts. Let's choose: $$u = \ln x \quad,\quad d v = x d x$$

Find du and v

To apply integration by parts, we need to find du and v. Differentiate u with respect to x to find du, and integrate dv to find v. $$ du = \frac{1}{x} dx $$ $$ v = \int x dx = \frac{1}{2}x^2 $$

Apply integration by parts formula

Now, we can apply the (definite) integration by parts formula: $$\int_{0}^{1} u d v = \Bigg[ uv \Bigg]_{0}^{1} - \int_{0}^{1} v d u$$ Substitute the values of u, v, and du we found in steps 1 and 2: $$\Bigg[\frac{1}{2}x^2\ln x \Bigg]_{0}^{1} - \int_{0}^{1} \frac{1}{2}x^2\frac{1}{x} d x$$

Simplify and evaluate the integrals

Now, let's simplify the expression: $$\Bigg[\frac{1}{2}x^2\ln x \Bigg]_{0}^{1} - \int_{0}^{1} \frac{1}{2}x d x$$ Evaluate the integral: $$\frac{1}{2}(1)^2\ln{(1)} - \frac{1}{2}(0)^2\ln{(0)} - \frac{1}{4}\Bigg[x^2\Bigg]_{0}^{1} = 0 - 0 - \frac{1}{4}(1-0) = -\frac{1}{4}$$ So the integral $$\int_{0}^{1} x \ln x d x$$ is equal to $$-\frac{1}{4}$$