Problem 31 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int(3 x+1)(4-x) d x$$

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the function \((3x+1)(4-x)\) with respect to x. Answer: The indefinite integral of the function \((3x+1)(4-x)\) with respect to x is \(6x^2 - x^3 + 4x - \frac{x^2}{2} + C\), where \(C\) is the constant of integration.

Step 1: Expand the function

In order to integrate \((3x+1)(4-x)\), we first need to expand it. Using the distributive property of multiplication, we get: $$(3x+1)(4-x) = 3x(4) - 3x(x) + 1(4) - 1(x) = 12x - 3x^2 + 4 - x.$$

Step 2: Integrate the expanded function

Now, we need to find the indefinite integral of the expanded function \(12x - 3x^2 + 4 - x\) with respect to x. To do this, we integrate each term separately: $$\int (12x - 3x^2 + 4 - x) dx = \int 12x dx - \int 3x^2 dx + \int 4 dx - \int x dx.$$ Using basic integration rules, we have: $$\int 12x dx = 12 \int x dx = 12\cdot \frac{x^2}{2} = 6x^2$$ $$\int 3x^2 dx = 3 \int x^2 dx = 3\cdot \frac{x^3}{3} = x^3$$ $$\int 4 dx = 4 \int dx = 4x$$ $$\int x dx = \frac{x^2}{2}.$$ Therefore, the indefinite integral of the given function is: $$\int (12x - 3x^2 + 4 - x) dx = 6x^2 - x^3 + 4x - \frac{x^2}{2} + C$$ where \(C\) is the constant of integration.

Step 3: Check the solution by differentiation

To check our integration, we take the derivative of the obtained function with respect to x: $$\frac{d}{dx} (6x^2 - x^3 + 4x - \frac{x^2}{2} + C)= 12x - 3x^2 + 4 - x.$$ Since the resulting derivative is the same as the initial function, our integration is correct.