Question

consider the function f(t)=7sec²(t) - 3t³. let f(t) be the antiderivative of f(t) with f(0)=0. then f(t) equals

Explanation:

Step1: Recall antiderivative rules

We know that the antiderivative of $\sec^{2}(t)$ is $\tan(t)$ and the antiderivative of $t^{n}$ is $\frac{t^{n + 1}}{n+1}+C$ for $n
eq - 1$.

Step2: Find antiderivative of $f(t)$

The antiderivative of $f(t)=7\sec^{2}(t)-3t^{3}$ is $F(t)=7\tan(t)-3\times\frac{t^{4}}{4}+C$.

Step3: Use the initial - condition

Since $F(0) = 0$, we substitute $t = 0$ into $F(t)$: $F(0)=7\tan(0)-\frac{3}{4}(0)^{4}+C$. Since $\tan(0)=0$, we have $0 = 0-0 + C$, so $C = 0$.

Answer:

$F(t)=7\tan(t)-\frac{3}{4}t^{4}$