Definite Integrals Question 8

Using the substitution $u=2x+1$ , $\displaystyle\int\limits_0^2 \sqrt{2x+1} dx$ is equivalent to

\displaystyle \frac{1}{2}\int\limits_{-1/2}^{1/2}\sqrt{u}   du

\displaystyle \frac{1}{2}\int\limits_0^2 \sqrt{u}   du

\displaystyle \frac{1}{2}\int\limits_1^5 \sqrt{u}   du

\displaystyle \int\limits_1^5 \sqrt{u}   du

Approach

u = 2x+1

du = 2   dx

dx = \frac{1}{2}  du

Upper Bound

x=2

u = 2(2)+1

u = 5

Lower Bound

x=0

u = 2(0)+1

u = 1

Substituting these values into the definite integral:

\int\limits_0^2 \sqrt{\colorbox{aqua}{$2x+1$}}   \colorbox{yellow}{$dx$}

\int\limits_1^5 \sqrt{\colorbox{aqua}{$u$}} \cdot \colorbox{yellow}{$\frac{1}{2}   du$}

= \boxed{\frac{1}{2}\int\limits_1^5 \sqrt{u}   du}