This problem involves finding an antiderivative through fundamental integration principles. To find an antiderivative of A'(t) = (1/2)t⁴ - 3t², we apply the power rule: ∫tⁿ dt = t^(n+1)/(n+1) + C. For (1/2)t⁴, we get (1/2)·t⁵/5 = (1/10)t⁵, and for -3t², we get -3·t³/3 = -t³. Thus, the antiderivative is (1/10)t⁵ - t³ + C. Choice D ((1/10)t⁵ - 3t² + C) incorrectly integrates the second term, forgetting to increase the exponent—a mistake showing confusion about the integration process. Always remember: integration reverses differentiation by increasing powers and dividing by the new power.

This problem requires finding a general antiderivative of A'(t) = 4t⁹ + 11, demonstrating basic antiderivative reasoning. To find the antiderivative, we apply the power rule for integration: ∫tⁿdt = t^(n+1)/(n+1) + C. For 4t⁹, we get 4t¹⁰/10 = (2/5)t¹⁰; for the constant 11, we get 11t. Combining these terms with the constant of integration gives (2/5)t¹⁰ + 11t + C. Choice B ((4/10)t¹⁰ + 11t + C) shows the unsimplified fraction 4/10 instead of the reduced form 2/5, though both are mathematically equivalent. The key integration strategy is to apply the power rule carefully and simplify fractions when possible for cleaner expressions.

This problem requires finding an antiderivative by applying the power rule for integration to each term. For M(x) = 12x⁵ + 3x², we integrate term by term: 12x⁵ becomes 12x⁶/6 = 2x⁶, and 3x² becomes 3x³/3 = x³. The antiderivative is 2x⁶ + x³ + C. Choice B (2x⁶ + 3x³ + C) incorrectly keeps the coefficient 3 when integrating 3x², failing to divide by the new exponent 3. Remember the integration formula: ∫axⁿ dx = axⁿ⁺¹/(n+1) + C, where you must divide by the new exponent to reverse the chain rule effect from differentiation.

This problem asks for a general antiderivative of f'(x) = (1/2)x⁷ - 6x, testing basic antiderivative reasoning skills. To find the antiderivative, we apply the power rule for integration to each term separately. For (1/2)x⁷, we get (1/2)·x⁸/8 = x⁸/16; for -6x, we get -6x²/2 = -3x². Including the constant of integration yields (1/16)x⁸ - 3x² + C. Choice B ((1/16)x⁸ - 6x² + C) incorrectly integrates -6x as -6x² instead of -3x², failing to divide by the new exponent 2. The transferable strategy is to remember that when integrating axⁿ, the result is a·x^(n+1)/(n+1), requiring division by the new exponent.

This problem requires finding an antiderivative using basic integration rules. To find an antiderivative of v(t) = 6t² - 4t + 9, we reverse the differentiation process by applying the power rule for integration: ∫tⁿ dt = tⁿ⁺¹/(n+1) + C. For 6t², we get 6t³/3 = 2t³; for -4t, we get -4t²/2 = -2t²; and for 9, we get 9t. Combining these terms gives 2t³ - 2t² + 9t + C. Choice E (2t³ - 2t² + 9 + C) incorrectly integrates the constant 9 as 9 instead of 9t, forgetting that the integral of a constant is that constant times the variable. Remember: to verify an antiderivative, differentiate your answer and check if you get back the original function.

This problem asks for a general antiderivative of g'(x) = 3x⁸ - 10x⁴ + 2, testing basic antiderivative reasoning skills. To find the antiderivative, we apply the power rule for integration to each term. For 3x⁸, we get 3x⁹/9 = (1/3)x⁹; for -10x⁴, we get -10x⁵/5 = -2x⁵; for the constant 2, we get 2x. Combining these with the constant of integration yields (1/3)x⁹ - 2x⁵ + 2x + C. Choice D ((1/9)x⁹ - 2x⁵ + 2x + C) incorrectly integrates 3x⁸ as (1/9)x⁹ instead of (1/3)x⁹, making an error in simplifying 3/9. The transferable strategy is to carefully simplify coefficients after applying the power rule: when integrating $ax^n$, the result is a/(n+1)·x^(n+1).

Finding antiderivatives requires careful application of the power rule, especially with fractional coefficients. To integrate v(t) = (3/2)t⁴ - 5t, we apply the rule to each term: (3/2)t⁴ becomes (3/2)t⁵/5 = (3/10)t⁵, and -5t becomes -5t²/2 = -(5/2)t². The antiderivative is (3/10)t⁵ - (5/2)t² + C. Choice E ((3/2)t⁵ - (5/2)t² + C) incorrectly keeps the original coefficient 3/2 when integrating (3/2)t⁴, forgetting to divide by the new exponent 5. When integrating terms with fractions, multiply the coefficient by 1/(n+1), which often creates more complex fractions that must be simplified.

This problem asks for a general antiderivative of the marginal cost function C'(x) = 9x² + 7x⁶, testing basic antiderivative reasoning skills. To find the antiderivative, we apply the power rule for integration: for xⁿ, the antiderivative is x^(n+1)/(n+1). For 9x², we get 9x³/3 = 3x³; for 7x⁶, we get 7x⁷/7 = x⁷. Including the constant of integration, the general antiderivative is 3x³ + x⁷ + C. Choice D (3x³ + x⁷) omits the crucial constant of integration C, which represents all possible vertical translations of the antiderivative function. The transferable strategy is to always include +C in indefinite integrals, as this constant captures the infinite family of functions that all have the same derivative.

This problem requires finding an antiderivative of the population growth rate P'(t) = 12t⁵ - 5t², demonstrating basic antiderivative reasoning. To find the antiderivative, we reverse the differentiation process using the power rule for integration. For 12t⁵, we get 12t⁶/6 = 2t⁶; for -5t², we get -5t³/3 = -5/3·t³. Combining these with the constant of integration gives 2t⁶ - (5/3)t³ + C. Choice B (2t⁶ - 5t³ + C) incorrectly integrates -5t² as -5t³ instead of -(5/3)t³, forgetting to divide by the new exponent. The key integration strategy is to carefully apply the power rule formula ∫tⁿdt = t^(n+1)/(n+1) + C, ensuring you divide by the new exponent (n+1).

This problem requires finding ∫p(t)dt where p(t) = 7t² + 14t⁵, demonstrating basic antiderivative reasoning through integration notation. To find the indefinite integral, we apply the power rule to each term: ∫tⁿdt = t^(n+1)/(n+1) + C. For 7t², we get 7t³/3 = (7/3)t³; for 14t⁵, we get 14t⁶/6 = (14/6)t⁶ = (7/3)t⁶. Including the constant of integration gives (7/3)t³ + (7/3)t⁶ + C. Choice B shows (14/6)t⁶ instead of the simplified (7/3)t⁶, though both are equivalent since 14/6 = 7/3. The key strategy is to simplify fractions in your final answer and always include +C for indefinite integrals.