Question

QUESTION 10 Find the most general antiderivative. $\int (8t^2 + \frac{1}{10}) dt$ $\frac{8}{3}t^3 + t + C$ $16t + \frac{1}{10} + C$ $24t^3 + \frac{1}{5}t^2 + C$ $\frac{8}{3}t^3 + \frac{1}{20}t^2 + C$

          QUESTION 10
Find the most general antiderivative.
$\int (8t^2 + \frac{1}{10}) dt$
$\frac{8}{3}t^3 + t + C$
$16t + \frac{1}{10} + C$
$24t^3 + \frac{1}{5}t^2 + C$
$\frac{8}{3}t^3 + \frac{1}{20}t^2 + C$

QUESTION 10
Find the most general antiderivative.
∫ (8t^2 + (1)/(10)) dt
(8)/(3)t^3 + t + C
16t + (1)/(10) + C
24t^3 + (1)/(5)t^2 + C
(8)/(3)t^3 + (1)/(20)t^2 + C

Added by Lisa G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Zhumagali Shomanov

09/02/2022

Step 1

Step 1: To find the antiderivative of the given function \int (8t^(2)+(t)/(10))dt, we need to integrate each term separately. Show more…

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QUESTION 10 Find the most general antiderivative. int (8t^(2)+(t)/(10))dt (8)/(3)t^(3)+t+C 16t+(1)/(10)+C 24t^(3)+(1)/(5)t^(2)+C (8)/(3)t^(3)+(t^(2))/(20)+C QUESTION 10 10 points Save Answer Find the most general antiderivative. J [g2+] dt O3+t+C O 16t + + C 10 O 24t3 +t2 + C O t3 +t2 + C 20