Answer
$\int^b_a=\begin{bmatrix}
e-1 & \frac{e^{2}-1}{2} & \frac{1}{3}\\
2(e-1) & 2(e^{2}-1) & \frac{5}{3}
\end{bmatrix}$
Work Step by Step
Given: $A(t)=\begin{bmatrix}
e^t & e^{2t} & t^2\\
2e^t & 4e^{2t} & 5t^2
\end{bmatrix}$
The antiderivative of the matrix function is given by:
$\int^b_aA(t)dt=\int^b_aa_{ij}(t)dt$
Hence here, $\int^b_a=\int ^1_0\begin{bmatrix}
e^t & e^{2t} & t^2\\
2e^t & 4e^{2t} & 5t^2
\end{bmatrix}=\begin{bmatrix}
e^t & \frac{e^{2}}{2} & \frac{1}{3}\\
2e & 2e^{2} & \frac{5}{3}
\end{bmatrix}-\begin{bmatrix}
1& \frac{1}{2} & 0\\
2& 2 &0
\end{bmatrix}=\begin{bmatrix}
e-1 & \frac{e^{2}-1}{2} & \frac{1}{3}\\
2(e-1) & 2(e^{2}-1) & \frac{5}{3}
\end{bmatrix}$
