Use the substitution formula to evaluate the integral.
-$$\int _ { 0 } ^ { 1 } \sqrt { x + 1 } d x$$
A) $3 \sqrt { 2 } - 3$
B) $\frac { 4 } { 3 } \sqrt { 2 }$
C) $2 \sqrt { 2 } - 1$
D) $\frac { 4 } { 3 } \sqrt { 2 } - \frac { 2 } { 3 }$

Question

Quizplus · Accepted Answer

Using the substitution method, let $u = x + 1$, then $du = dx$. The limits of integration change accordingly: when $x = 0$, $u = 1$, and when $x = 1$, $u = 2$. The integral becomes $\int _ { 1 } ^ { 2 } \sqrt { u } du$, which evaluates to $\frac { 2 } { 3 } u ^ { \frac { 3 } { 2 } }$ evaluated from 1 to 2. This results in $\frac { 2 } { 3 } (2 ^ { \frac { 3 } { 2 } } - 1 ^ { \frac { 3 } { 2 } }) = \frac { 4 } { 3 } \sqrt { 2 } - \frac { 2 } { 3 }$.

Use the Substitution Formula to Evaluate the Integral ∫01x+1dx\int _ { 0 } ^ { 1 } \sqrt { x + 1 } d x∫01​x+1​dx

Use the Substitution Formula to Evaluate the Integral $\int _ { 0 } ^ { 1 } \sqrt { x + 1 } d x$