consider the function $f(t) = 6 \sec^2(t) - 8t^3$.

let $f(t)$ be the antiderivative of $f(t)$ with $f(0) = 0$. then
$f(t) =$

Question

consider the function $f(t) = 6 \sec^2(t) - 8t^3$.

let $f(t)$ be the antiderivative of $f(t)$ with $f(0) = 0$. then
$f(t) =$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Antiderivatives", "Power Rule for Integration" ], "new_concepts": [], "current_concepts": [ "Antiderivatives", "Power Rule for Integration", "Trigonometric Antiderivatives" ] } ## Find the general antiderivative $$ F(t) = \int (6\sec^2(t) - 8t^3) \, dt = 6\tan(t) - 2t^4 + C $$ ## Solve for the constant of integration $$ \begin{aligned} F(0) &= 6\tan(0) - 2(0)^4 + C = 0 \ 0 - 0 + C &= 0 \ C &= 0 \end{aligned} $$ ## State the final particular antiderivative $$ F(t) = 6\tan(t) - 2t^4 $$ Consider the function $f(t) = 6\sec^2(t) - 8t^3$. Let $F(t)$ be the antiderivative of $f(t)$ with $F(0) = 0$. Then $F(t) =$ $6\tan(t) - 2t^4$ { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Antiderivatives" ] }

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QUESTION IMAGE

Question

Explanation:

Find the general antiderivative

Solve for the constant of integration

State the final particular antiderivative

Answer: