PUNE MCA Exam 2014 Question Paper with answers for online practice

Concept:The integral of a matrix is computed element-wise, and for odd functions over symmetric limits, the integral is zero.Explanation:Given $f(x) = \begin{bmatrix} x^3 & \sin x \\ 1 & 2 \end{bmatrix}$, we need $\int_{-a}^{a} f(x) \, dx$.The integral is performed on each element:• $\int_{-a}^{a} x^3 \, dx = 0$ (since $x^3$ is odd).• $\int_{-a}^{a} \sin x \, dx = 0$ (since $\sin x$ is odd).• $\int_{-a}^{a} 1 \, dx = 2a$.• $\int_{-a}^{a} 2 \, dx = 4a$.Thus, $\int_{-a}^{a} f(x) \, dx = \begin{bmatrix} 0 & 0 \\ 2a & 4a \end{bmatrix}$, a $2\times 2$ matrix. However, the given options are scalars. The most reasonable interpretation is that the question asks for the integral of the determinant of $f(x)$ (or the determinant of the integrated matrix).The determinant of $f(x)$ is $\det(f(x)) = x^3 \cdot 2 - \sin x \cdot 1 = 2x^3 - \sin x$, an odd function. Hence, $\int_{-a}^{a} (2x^3 - \sin x) \, dx = 0$.Answer:0