consider the function $f(t)=10sec^{2}(t)-8t^{3}$. let $f(t)$ be the antiderivative of $f(t)$ with $f(0) =…

Explanation:

Step1: Find antiderivative of each term

The antiderivative of $10\sec^{2}(t)$ is $10\tan(t)$ (since the derivative of $\tan(t)$ is $\sec^{2}(t)$) and the antiderivative of $- 8t^{3}$ is $-8\times\frac{t^{4}}{4}=-2t^{4}$ (using the power - rule for integration $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C$, $n\neq - 1$). So the general antiderivative of $f(t)$ is $F(t)=10\tan(t)-2t^{4}+C$.

Step2: Use the initial condition

We know that $F(0) = 0$. Substitute $t = 0$ into $F(t)$: $F(0)=10\tan(0)-2\times0^{4}+C$. Since $\tan(0)=0$, we have $0 = 0-0 + C$, so $C = 0$.

Answer:

$10\tan(t)-2t^{4}$