Answer to Question #174615 in Calculus for Phyroe

Question

Evaluate the integral of √(3x + 1) dx from 1 to 5

Answer 1

Solution.

$$\int\limits_1^5\sqrt{3x+1}dx.$$

Make a substitution:

$t=3x+1,$ then $dt=3dx,$ hence $dx=\frac{1}{3}dt.$

$\int\sqrt{3x+1}dx=\int\frac{1}{3}\sqrt{t}dt=\newline =\frac{1}{3}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2}{9}t^{\frac{3}{2}}=\newline =\frac{2}{9}(3x+1)^{\frac{3}{2}}.$

So,

$\int\limits_1^5\sqrt{3x+1}dx=(\frac{2}{9}(3x+1)^{\frac{3}{2}})|_1^5=\newline =\frac{128}{9}-\frac{16}{9}=\frac{112}{9}=12\frac{4}{9}.$

Answer: $\int\limits_1^5\sqrt{3x+1}dx=12\frac{4}{9}.$