Evaluate the following definite integral. $\int _ { 1 } ^ { 4 } \frac { 1 } { \sqrt { 6 z + 7 } } d z$ Use a graphing utility to check your answer.
A) $\frac { \sqrt { 31 } - \sqrt { 13 } } { 6 }$
B) $\frac { \sqrt { 31 } - \sqrt { 13 } } { 3 }$
C) $\frac { \sqrt { 31 } + \sqrt { 13 } } { 3 }$
D) $\frac { \sqrt { 31 } + \sqrt { 13 } } { 6 }$
E) $\frac { \sqrt { 13 } - \sqrt { 31 } } { 3 }$

Question

Quizplus · Accepted Answer

The integral can be solved using substitution. Let $ u = 6z + 7 $, then $ du = 6 \, dz $ or $ dz = \frac{1}{6} \, du $. The limits change from $ z = 1 $ to $ u = 13 $ and $ z = 4 $ to $ u = 31 $. The integral becomes $\int_{13}^{31} \frac{1}{\sqrt{u}} \cdot \frac{1}{6} \, du = \frac{1}{6} \int_{13}^{31} u^{-1/2} \, du$. This evaluates to $\frac{1}{6} \left[ 2u^{1/2} \right]_{13}^{31} = \frac{1}{3} \left( \sqrt{31} - \sqrt{13} \right)$.

Evaluate the Following Definite Integral ∫1416z+7dz\int _ { 1 } ^ { 4 } \frac { 1 } { \sqrt { 6 z + 7 } } d z∫14​6z+7​1​dz

Evaluate the Following Definite Integral $\int _ { 1 } ^ { 4 } \frac { 1 } { \sqrt { 6 z + 7 } } d z$