8 (a) Given that $u = 2^x$, write down an expression for \( \frac{du}{dx} \) 8 (b) Find the exact value of \( \int_{2}^{3} \sqrt{3 + 2x} \, dx \) Fully justify your answer. - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 1

To solve the integral ( I = \int_{2}^{3} \sqrt{3 + 2x} , dx ), we start by simplifying the integrand. Let:

1. Substitution:

   Let ( u = 3 + 2x ) Thus, ( \frac{du}{dx} = 2 ) or ( dx = \frac{du}{2} )

   Changing the limits:

   • When ( x = 2, \ u = 3 + 4 = 7 )
   • When ( x = 3, \ u = 3 + 6 = 9 )

   Now substituting: $I = \int_{7}^{9} \sqrt{u} \frac{du}{2}$

   2. Integrate:

   We have: $I = \frac{1}{2} \int_{7}^{9} u^{1/2} \ du \ = \frac{1}{2} \left[ \frac{u^{3/2}}{\frac{3}{2}} \right]_{7}^{9} = \frac{1}{3} \left[ u^{3/2} \right]_{7}^{9}$

   Evaluating at the limits: $I = \frac{1}{3} \left[ 9^{3/2} - 7^{3/2} \right]$ Since ( 9^{3/2} = 27 ) and ( 7^{3/2} = 7 \sqrt{7} $$:

   $I = \frac{1}{3} \left[ 27 - 7\sqrt{7} \right]$ Hence, the value of the integral is: $I = \frac{27 - 7\sqrt{7}}{3}$