If the accumulation function $F ( x ) \text { is given by } F ( x ) = \int _ { 0 } ^ { x } \left( \frac { 1 } { 11 } t ^ { 2 } + 5 \right) d t \text {, evaluate }$ $F( 9 )$
A) $A = \frac { 738 } { 11 }$
B) $A = \frac { 356 } { 33 }$
C) $A = \frac { 852 } { 11 }$
D) $A = \frac { 298 } { 11 }$
E) $A = \frac { 1068 } { 11 }$

Question

Quizplus · Accepted Answer

To evaluate $F(9)$, integrate the function $\frac{1}{11}t^2 + 5$ from 0 to 9. The integral is $\frac{1}{11} \cdot \frac{t^3}{3} + 5t$ evaluated from 0 to 9, which simplifies to $\frac{1}{33} \cdot 9^3 + 5 \cdot 9 = \frac{729}{33} + 45 = \frac{729}{33} + \frac{1485}{33} = \frac{2214}{33} = \frac{738}{11}$.

If the Accumulation Function F(x) is given by F(x)=∫0x(111t2+5)dt, evaluate F ( x ) \text { is given by } F ( x ) = \int _ { 0 } ^ { x } \left( \frac { 1 } { 11 } t ^ { 2 } + 5 \right) d t \text {, evaluate }F(x) is given by F(x)=∫0x​(111​t2+5)dt, evaluate

If the Accumulation Function $F ( x ) \text { is given by } F ( x ) = \int _ { 0 } ^ { x } \left( \frac { 1 } { 11 } t ^ { 2 } + 5 \right) d t \text {, evaluate }$