Kerala KEAM 5 June 2024 Solved Paper

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). Quick Tip: 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.