Answer
$\int^b_aA(t)dt=\begin{bmatrix}
\frac{e^{2}-1 }{2}& -\frac{1-\cos 2}{2} \\
-\frac{14}{3} & 1 \\
\tan 1 & \frac{1}{2}+\cos 1
\end{bmatrix}$
Work Step by Step
Given: $A(t)=\begin{bmatrix}
e^{2t} & \sin^{2t} \\
t^2 -5 & te^{t} \\
\sec^2t & 3t-\sin t
\end{bmatrix}$
The antiderivative of the matrix function is given by:
$\int^b_aA(t)dt=\int^b_aa_{ij}(t)dt$
Hence here, $\int^b_a=\int ^1_0\begin{bmatrix}
e^{2t} & \sin^{2t} \\
t^2 -5 & te^{t} \\
\sec^2t & 3t-\sin t
\end{bmatrix}=\begin{bmatrix}
\frac{e^{2} }{2}& -\frac{\cos 2}{2} \\
-\frac{14}{3} & 0 \\
\tan 1 & \frac{3}{2}+\cos 1
\end{bmatrix}-\begin{bmatrix}
\frac{1}{2}& -\frac{-1}{2} \\
0 &-1 \\
0 & 1
\end{bmatrix}=\begin{bmatrix}
\frac{e^{2}-1 }{2}& -\frac{1-\cos 2}{2} \\
-\frac{14}{3} & 1 \\
\tan 1 & \frac{1}{2}+\cos 1
\end{bmatrix}$
