Evaluate $$\int_{4}^{9} \frac{1}{\sqrt{(2x+9)}}dx.$$ - Scottish Highers Maths - Question 14 - 2018

We can rewrite the integrand as: $\int (2x+9)^{-1/2}dx.$

Using the power rule for integration, we have: $\int (2x+9)^{-1/2}dx = \frac{(2x+9)^{1/2}}{(1/2)}\cdot \frac{1}{2} = (2x+9)^{1/2} + C.$

Now we apply the limits of integration from 4 to 9: $\left[ (2(9)+9)^{1/2} - (2(4)+9)^{1/2} \right].$

Calculating gives us:

1. Evaluate at the upper limit:

   • $(2(9)+9)^{1/2} = (18+9)^{1/2} = 27^{1/2} = 3\sqrt{3}$

2. Evaluate at the lower limit:

   • $(2(4)+9)^{1/2} = (8+9)^{1/2} = 17^{1/2} = \sqrt{17}$

Combining both results gives: $3\sqrt{3} - \sqrt{17}.$