Question

The value of \( \int_{-4}^{-2} \left[ (x+3)^3 + 2 + (x+3)\cos(x+3) \right] \, dx \) is equal to:

• A. 3
• B. -2
• C. -1
• D. 1
• E. 4

Hint:

When dealing with odd and even functions, remember that the integral of an odd function over a symmetric interval is zero. This can simplify your calculations.

Answer 1

Answer: (E)

We are given the integral: \[ I = \int_{-4}^{-2} \left[ (x+3)^3 + 2 + (x+3)\cos(x+3) \right] \, dx \] To simplify this, let's make a substitution. Let: \[ u = x + 3 \] Thus: \[ du = dx \] Also, the limits of integration change accordingly. 
When \( x = -4 \), \( u = -1 \), and when \( x = -2 \), \( u = 1 \). 
Now, substitute into the integral: \[ I = \int_{-1}^{1} \left[ u^3 + 2 + u \cos u \right] \, du \] 
We can break the integral into three parts: \[ I = \int_{-1}^{1} u^3 \, du + \int_{-1}^{1} 2 \, du + \int_{-1}^{1} u \cos u \, du \] 
Now, evaluate each part: 
1. The integral of \( u^3 \): \[ \int_{-1}^{1} u^3 \, du = \left[ \frac{u^4}{4} \right]_{-1}^{1} = \frac{1^4}{4} - \frac{(-1)^4}{4} = \frac{1}{4} - \frac{1}{4} = 0 \] 
2. The integral of \( 2 \): \[ \int_{-1}^{1} 2 \, du = 2 \times (1 - (-1)) = 2 \times 2 = 4 \] 
3. The integral of \( u \cos u \): The function \( u \cos u \) is odd because \( u \) is odd and \( \cos u \) is even. 
The integral of an odd function over a symmetric interval (from -1 to 1) is zero: \[ \int_{-1}^{1} u \cos u \, du = 0 \] 
Now, add the results of the three integrals: \[ I = 0 + 4 + 0 = 4 \] 
Thus, the value of the integral is 4, which corresponds to option (E).