MHTCET 15 May 2024 Shift 1 Paper

Concept:Rewrite $x^{3} + 3x^{2} + 3x + 5$ as $(x+1)^{3} + 4$, then substitute $t = x+1$ and apply the property of odd functions over symmetric limits.Explanation:Consider the given integral:$$I = \int_{-2}^{0} \left[x^{3} + 3x^{2} + 3x + 5 + (x+1)\cos(x+1)\right] dx$$Observe that the cubic expression can be written as:$$x^{3} + 3x^{2} + 3x + 5 = (x+1)^{3} + 4$$So the integral becomes:$$I = \int_{-2}^{0} \left[(x+1)^{3} + 4 + (x+1)\cos(x+1)\right] dx$$Put $t = x+1$, which gives $dx = dt$.When $x = -2$, $t = -1$; and when $x = 0$, $t = 1$.Hence:$$I = \int_{-1}^{1} \left(t^{3} + 4 + t\cos t\right) dt$$Here, $t^{3}$ is an odd function and $t\cos t$ is also an odd function.The integral of any odd function over $[-1, 1]$ is zero, so both odd terms drop out.Therefore, only the constant term remains:$$I = \int_{-1}^{1} 4\, dt = 4\left[t\right]_{-1}^{1} = 4\left(1 - (-1)\right) = 8$$Answer:The value of the integral is $8$.Since this value is not listed among the options, the provided options appear to be incorrect.