$\int _ { 0 } ^ { 8 } \left( \sqrt [ 3 ] { t ^ { 2 } } + \frac { 1 } { \sqrt { t + 1 } } \right) d t = \frac { 116 } { 5 }$

Name: $\int _ { 0 } ^ { 8 } \left( \sqrt [ 3 ] { t ^ { 2 } } + \frac { 1 } { \sqrt { t + 1 } } \right) d t = \frac { 116 } { 5 }$
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Question

Quizplus · Accepted Answer

The integral of $\sqrt [ 3 ] { t ^ { 2 } } + \frac { 1 } { \sqrt { t + 1 } } $ from 0 to 8 is $\frac { 116 } { 5 }$. This can be verified by using the power rule and the chain rule to integrate each term separately.

∫08(t23+1t+1)dt=1165\int _ { 0 } ^ { 8 } \left( \sqrt [ 3 ] { t ^ { 2 } } + \frac { 1 } { \sqrt { t + 1 } } \right) d t = \frac { 116 } { 5 }∫08​(3t2​+t+1​1​)dt=5116​

$\int _ { 0 } ^ { 8 } \left( \sqrt [ 3 ] { t ^ { 2 } } + \frac { 1 } { \sqrt { t + 1 } } \right) d t = \frac { 116 } { 5 }$