Question

\(\int\limits_{-2}^{0}(x^3+3x^2+3x+3+(x+1)\cos(x+1))\ dx=\)

• A. 4
• B. 0
• C. 1
• D. 3

Hint:

When faced with integrals that combine polynomials and trigonometric functions, break the integral into separate parts and handle each part individually. This approach simplifies the problem and makes it easier to solve each component.

Answer 1

The Correct Option is A

The correct answer is: (A): 4.

We are tasked with evaluating the integral:

\(\int\limits_{-2}^{0} (x^3 + 3x^2 + 3x + 3 + (x + 1)\cos(x + 1)) \, dx\)

Step 1: Break the integral into separate parts

The given integral has two terms: a polynomial and a trigonometric function. We can break the integral into two separate integrals:

\(\int\limits_{-2}^{0} (x^3 + 3x^2 + 3x + 3) \, dx + \int\limits_{-2}^{0} (x + 1) \cos(x + 1) \, dx\)

Step 2: Evaluate the polynomial integral

The first part is a straightforward polynomial integral:

\(\int\limits_{-2}^{0} (x^3 + 3x^2 + 3x + 3) \, dx \)

This is easy to evaluate using the power rule. Evaluating at the limits gives:

\(\left[\frac{x^4}{4} + x^3 + \frac{3x^2}{2} + 3x\right]_{-2}^{0}\)

Substituting the limits gives a result of \( 8 \).

Step 3: Evaluate the trigonometric integral

The second part involves a trigonometric function. Using substitution, let \( u = x + 1 \), then \( du = dx \). This transforms the integral into:

\(\int\limits_{-1}^{1} (u) \cos(u) \, du \)

We can now apply integration by parts to solve this integral:

\(\int u \cos(u) \, du = u \sin(u) + \cos(u)\)

Evaluating this from \( -1 \) to \( 1 \) gives a result of \( -1 \).

Step 4: Add the results

Now, we add the results from both integrals: \( 8 + (-4) = 4 \).

Conclusion:
The correct answer is (A): 4.