antiderivatives

Question

Consider the function 𝑓(𝑡)=8sec2(𝑡)-9𝑡3. Let 𝐹(𝑡) be the antiderivative of 𝑓(𝑡) with 𝐹(0)=0.

Then 𝐹(3)=   

Answer 1

The trick with this question is to understand how the constant "C" term works with integration. The F(0)=... is an example of an initial condition, which makes it possible to solve for C simply by inputting 0 into the t's of the anti-derivative. Using this cC value, you are able to solve for F(3) by plugging 3 into the anti-derivative. The process would go as follows:

f(t) = 8sec²(t) - 9t³

F(t) = 8tan(t) - 9/4*x⁴ + C

F(0) = 0; therefore 0 = 8tan(0) - 9/4*0⁴ + C

tan(0) = 0 and 0⁴ = 0, leaving you with 0 = C.

With this, we can plug 3 into the antiderivative and solve.

F(3) = 8tan(3) - 9/4*3⁴ + 0

F(3) = -1.14 + 9*81/4 = 181.1

Notes: I've assumed the question is using radians. As for the integration, it is done using a memorized rule for the sec²() to tan(), and the t³ is integrated using the inverse power rule.