Answer
$f(x)=\displaystyle \frac{9}{28}t^{7/3}+\cos t+(\frac{19}{28}-\cos 1)t+1$
Work Step by Step
$f''(t)=t^{1/3}-\cos t$
Using the antiderivatives table,
$f'(x)=\displaystyle \frac{t^{4/3}}{4/3}-\sin t+C=\frac{3}{4}t^{4/3}-\sin t+C$
Using the antiderivatives table,
$f(x)=\displaystyle \frac{3}{4}(\frac{t^{7/3}}{7/3})+\cos t+Ct+D$
$f(x)=\displaystyle \frac{9}{28}t^{7/3}+\cos t+Ct+D$
$\left\{\begin{array}{llll}
f(0)=2 & \Rightarrow & 0+1+0+D=2 & \\
& & D=1 & \\
& & & \\
f(1)=2 & \Rightarrow & & \frac{9}{28}+\cos 1+C+1=2\\
& & & C=\frac{19}{28}-\cos 1\\
& & &
\end{array}\right.$
$f(x)=\displaystyle \frac{9}{28}t^{7/3}+\cos t+(\frac{19}{28}-\cos 1)t+1$
