Question

assignment 10: problem 6 (1 point) consider the function f(t) = 6 sec²(t) - 2t³. let f(t) be the antiderivative of f(t) with f(0) = 0. then f(0.8) = you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor preview my answers submit answers

Explanation:

Step1: Find the antiderivative of \( f(t) \)

The function is \( f(t) = 6\sec^2(t) - 2t^3 \). The antiderivative of \( \sec^2(t) \) is \( \tan(t) \), and the antiderivative of \( t^n \) is \( \frac{t^{n + 1}}{n+1} \) (for \( n
eq - 1 \)). So, the antiderivative \( F(t) \) of \( f(t) \) is:
\( F(t)=6\tan(t)-2\times\frac{t^{4}}{4}+C = 6\tan(t)-\frac{1}{2}t^{4}+C \)

Step2: Determine the constant \( C \)

We know that \( F(0) = 0 \). Substitute \( t = 0 \) into \( F(t) \):
\( F(0)=6\tan(0)-\frac{1}{2}(0)^{4}+C \)
Since \( \tan(0)=0 \), we have \( 0 = 0 - 0 + C \), so \( C = 0 \). Thus, \( F(t)=6\tan(t)-\frac{1}{2}t^{4} \)

Step3: Calculate \( F(0.8) \)

Substitute \( t = 0.8 \) into \( F(t) \):
First, calculate \( \tan(0.8) \approx 1.0296 \) (using a calculator). Then, \( t^{4}=(0.8)^{4}=0.4096 \)
\( F(0.8)=6\times1.0296-\frac{1}{2}\times0.4096 \)
\( F(0.8)=6.1776 - 0.2048 \)
\( F(0.8)=5.9728 \) (approximate value, more accurately, using calculator for precise computation:
\( 6\tan(0.8)-\frac{1}{2}(0.8)^4 \)
\( \tan(0.8)\approx1.029619417 \)
\( 6\times1.029619417 = 6.177716502 \)
\( \frac{1}{2}(0.8)^4=\frac{1}{2}\times0.4096 = 0.2048 \)
\( F(0.8)=6.177716502-0.2048 = 5.972916502\approx5.97 \) (or more precise value depending on calculator precision))

Answer:

\( \approx5.97 \) (the more precise value is approximately \( 5.9729 \))