Question

$\text{Let}f\left(x\right)=\left|\begin{array}{ccc}a\left(x\right)& b\left(x\right)& c\left(x\right)\\ m\left(x\right)& n\left(x\right)& l\left(x\right)\\ g\left(x\right)& h\left(x\right)& k\left(x\right)\end{array}\right|\text{then}$

${\int }_0^{a}f\left(x\right)\text{is}\left|\begin{array}{ccc}{\int }_0^{a}a\left(x\right)& {\int }_0^{a}b\left(x\right)& {\int }_0^{a}c\left(x\right)\\ {\int }_0^{a}m\left(x\right)& {\int }_0^{a}n\left(x\right)& {\int }_0^{a}l\left(x\right)\\ {\int }_0^{a}g\left(x\right)& {\int }_0^{a}h\left(x\right)& {\int }_0^{a}k\left(x\right)\end{array}\right|$

• A. False
• B. True

Answer 1

The correct option is A

False

Integration in determinants is not straight forward. But is is simple in some special cases like only one of the rows having elements as function of x and we will discuss only such cases in this standard.

Suppose $f\left(x\right)=\left|\begin{array}{ccc}a\left(x\right)& b\left(x\right)& c\left(x\right)\\ 4& 3& 4\\ 1& 5& 3\end{array}\right|$ , then ${\int }_0^{a}f\left(x\right)$ will be
$\left|\begin{array}{ccc}{\int }_0^{a}a\left(x\right)& {\int }_0^{a}b\left(x\right)& {\int }_0^{a}c\left(x\right)\\ 4& 3& 7\\ 1& 5& 3\end{array}\right|$, but if

$f\left(x\right)=\left|\begin{array}{ccc}a\left(x\right)& b\left(x\right)& c\left(x\right)\\ m\left(x\right)& n\left(x\right)& l\left(x\right)\\ g\left(x\right)& h\left(x\right)& k\left(x\right)\end{array}\right|$,
then ${\int }_0^{a}f\left(x\right)$ is not equal to

$\left|\begin{array}{ccc}{\int }_0^{a}a\left(x\right)& {\int }_0^{a}b\left(x\right)& {\int }_0^{a}c\left(x\right)\\ {\int }_0^{a}m\left(x\right)& {\int }_0^{a}n\left(x\right)& {\int }_0^{a}l\left(x\right)\\ {\int }_0^{a}g\left(x\right)& {\int }_0^{a}h\left(x\right)& {\int }_0^{a}k\left(x\right)\end{array}\right|$

There should be only one row with functions of x. Only then we can apply integration to determinants in the way we discussed. Else expand the determinant and then integrate the function