How do you find the definite integral of #(sqrt(2x+1))dx# from #[1, 4]#?

Question

Luke Alvarez · Accepted Answer

It is #int sqrt(2x+1)dx=1/3*int (2x+1)'*sqrt(2x+1)dx=1/3*[2x+1]^(3/2)+c# Hence #int_1^4 sqrt(2x+1)dx=[1/3*[2x+1]^(3/2)]_1^4=9-sqrt(3) ~~ 7.2679#

Aurora Alvarado · Answer

undefined To find the definite integral of $ \sqrt{2x+1} $ from $ x = 1 $ to $ x = 4 $, you can use the definite integral formula:

$$ \int_{1}^{4} \sqrt{2x+1} \, dx $$

1. First, find the antiderivative of $ \sqrt{2x+1} $ with respect to $ x $.
2. Evaluate the antiderivative at the upper limit of integration (4) and subtract the value of the antiderivative at the lower limit of integration (1).

$$ \left[ \frac{2}{3}(2x+1)^{\frac{3}{2}} \right]_{1}^{4} $$

3. Plug in the upper limit (4) and lower limit (1) into the antiderivative formula and subtract.

$$ = \left( \frac{2}{3}(2(4)+1)^{\frac{3}{2}} \right) - \left( \frac{2}{3}(2(1)+1)^{\frac{3}{2}} \right) $$

4. Calculate the values inside the parentheses.

$$ = \left( \frac{2}{3}(9)^{\frac{3}{2}} \right) - \left( \frac{2}{3}(3)^{\frac{3}{2}} \right) $$

5. Simplify and compute.

$$ = \left( \frac{2}{3}(27) \right) - \left( \frac{2}{3}(3\sqrt{3}) \right) $$

$$ = \left( \frac{54}{3} \right) - \left( \frac{6\sqrt{3}}{3} \right) $$

$$ = 18 - 2\sqrt{3} $$

So, the definite integral of $ \sqrt{2x+1} $ from $ x = 1 $ to $ x = 4 $ is $ 18 - 2\sqrt{3} $.