Problem 6
Question
$$ \int(\ln x)^{2} d x $$
Step-by-Step Solution
Verified Answer
The integral is \( x((\ln x)^2 - 2 \ln x + 2) + C \).
1Step 1: Use Integration by Parts Formula
Recall the formula for integration by parts: \[\int u dv = uv - \int v du\]Choose \(u = (\ln x)^2\) and \(dv = dx\).
2Step 2: Differentiate and Integrate
Differentiate \(u\):\[du = 2 \ln(x) \cdot \frac{1}{x} dx = \frac{2 \ln(x)}{x} dx\]Integrate \(dv\):\[v = \int dx = x\]
3Step 3: Apply the Integration by Parts
Substitute \(u, v, du,on\) back into the integration by parts formula:\[\int (\ln x)^2 dx = x(\ln x)^2 - \int x \cdot \frac{2 \ln x}{x} dx\]This simplifies to:\[\int (\ln x)^2 dx = x(\ln x)^2 - 2 \int \ln x \, dx\]
4Step 4: Integrate \(\int \ln x \, dx\)
Use integration by parts again, letting \(u = \ln x\) and \(dv = dx\). Thus, \(du = \frac{1}{x} dx\) and \(v = x\):\[\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - \int dx = x \ln x - x + C\]
5Step 5: Combine the Results
Substitute back into the expression from Step 3:\[\int (\ln x)^2 dx = x(\ln x)^2 - 2(x \ln x - x)\]Simplify it:\[\int (\ln x)^2 dx = x(\ln x)^2 - 2x \ln x + 2x + C\]Combine like terms:\[\int (\ln x)^2 dx = x((\ln x)^2 - 2 \ln x + 2) + C\]
Key Concepts
calculus for business and economics
calculus for business and economics
Calculus plays a crucial role in business and economics, helping in optimization, marginal analysis, and understanding changes over time. Among the integration techniques, definite integrals are particularly powerful:
For example, if a business wants to forecast profits from sales, they might use definite integral to calculate the total profit over a particular period, given by: \[\text{Total Profit} = \int_a^b \text{Profit \, Function} \, dx\]Utilizing integration techniques, businesses can make informed decisions by quantifying various continuous changes effectively.
- Optimization: Businesses use calculus to find the best conditions, like maximizing profit or minimizing costs. This generally involves finding where the derivative (rate of change) is zero.
- Marginal Analysis: The derivative of a cost function, revenue function, or profit function helps determine the additional cost or revenue for making one more unit.
- Consumer and Producer Surplus: Areas under demand and supply curves can be calculated using definite integrals, allowing businesses to assess economic welfare:
For example, if a business wants to forecast profits from sales, they might use definite integral to calculate the total profit over a particular period, given by: \[\text{Total Profit} = \int_a^b \text{Profit \, Function} \, dx\]Utilizing integration techniques, businesses can make informed decisions by quantifying various continuous changes effectively.
Other exercises in this chapter
Problem 4
$$ \int_{-9}^{-1} \frac{y d y}{\sqrt{4-5 y}} $$
View solution Problem 5
$$ \int_{1}^{4} \frac{\ln \sqrt{s}}{\sqrt{s}} d s $$
View solution Problem 7
$$ \int_{-2}^{1}(2 x+1)(x+3)^{3 / 2} d x $$
View solution Problem 8
$$ \int \frac{w^{3}}{\sqrt{1+w^{2}}} d w $$
View solution