Question

Evaluate \[ I=\int_{-2}^{0} \Big(x^3+3x^2+3x+3+(x+1)\cos(x+1)\Big)\,dx \]

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(1\) \bigskip

Hint:

For definite integrals over symmetric limits \([-a,a]\): \[ \int_{-a}^{a}\text{(odd function)}\,dx=0. \] After the substitution \(t=x+1\), the odd terms vanish immediately, leaving only the constant term.

Answer 1

The Correct Option is A

Step 1: Recognize a useful substitution. Let \[ t=x+1. \] Then \[ x^3+3x^2+3x+1=(x+1)^3=t^3. \] Hence \[ x^3+3x^2+3x+3 = (x+1)^3+2 = t^3+2. \] Therefore, \[ I = \int_{-2}^{0} \Big((x+1)^3+2+(x+1)\cos(x+1)\Big)\,dx. \] Using \(t=x+1\), \[ dx=dt, \] and limits become \[ x=-2 \Rightarrow t=-1, \qquad x=0 \Rightarrow t=1. \] Thus \[ I = \int_{-1}^{1} \Big(t^3+t\cos t+2\Big)\,dt. \]

Step 2: Use symmetry. Observe: \[ t^3 \] is an odd function, \[ t\cos t \] is also odd (odd \(\times\) even). Hence \[ \int_{-1}^{1} t^3\,dt=0, \] \[ \int_{-1}^{1} t\cos t\,dt=0. \] Therefore, \[ I = \int_{-1}^{1}2\,dt. \] \[ = 2[t]_{-1}^{1}. \] \[ = 2(1-(-1)). \] \[ =4. \]

Step 3: Write the final answer. \[ \boxed{4} \]