8 (a) Given that $u = 2^x$, write down an expression for \( \frac{du}{dx} \) 8 (b) Find the exact value of \( \int_0^2 \sqrt{3 + 2x} \, dx \) Fully justify your answer. - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 1

To solve ( \int_0^2 \sqrt{3 + 2x} , dx ), we will use substitution.

Step 1: Substitution

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

When ( x = 0 ), ( u = 3 ) and when ( x = 2 ), ( u = 7 ).

Step 2: Integral transformation

The integral transforms as follows:

$I = \int_0^2 \sqrt{3 + 2x} \, dx = \int_{3}^{7} \sqrt{u} \cdot \frac{du}{2} = \frac{1}{2} \int_{3}^{7} u^{1/2} \, du$

Step 3: Integrate

Carrying out the integration:

$= \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_{3}^{7} = \frac{1}{3} \left[ u^{3/2} \right]_{3}^{7}$

Step 4: Substitute limits

Now substituting the limits:

$= \frac{1}{3} \left[ 7^{3/2} - 3^{3/2} \right] = \frac{1}{3} \left[ 7 \sqrt{7} - 3 \sqrt{3} \right]$

Thus, the exact value of the integral is:

$\frac{1}{3} \left[ 7 \sqrt{7} - 3 \sqrt{3} \right]$