Use integration to find $$\int_{1}^{4} \left( \frac{x^3}{6} + \frac{1}{3x^2} \right) dx$$ giving your answer in the form $a + b\sqrt{3}$, where $a$ and $b$ are constants to be determined. - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 1

The first integral is:

$\int \frac{x^3}{6} dx = \frac{1}{6} \cdot \frac{x^4}{4} = \frac{x^4}{24}$

Evaluating this from 1 to 4:

$\left[ \frac{4^4}{24} - \frac{1^4}{24} \right] = \left[ \frac{256}{24} - \frac{1}{24} \right] = \frac{255}{24}$

The second integral is:

$\int \frac{1}{3x^2} dx = -\frac{1}{3} \cdot \frac{1}{x}$

Evaluating this from 1 to 4:

$\left[ -\frac{1}{3 \cdot 4} - \left( -\frac{1}{3 \cdot 1} \right) \right] = \left[ -\frac{1}{12} + \frac{1}{3} \right] = \left[ -\frac{1}{12} + \frac{4}{12} \right] = \frac{3}{12} = \frac{1}{4}$

So, adding the results from both integrals gives:

$I = \frac{255}{24} + \frac{1}{4}$

Converting \frac{1}{4} = \frac{6}{24}$, we find:

$I = \frac{255}{24} + \frac{6}{24} = \frac{261}{24}$

We need to express the result in the form $a + b\sqrt{3}$. To achieve this, we can approximatively express:

$\frac{261}{24} = 10.875$

We can thus represent it as:

$10 + 0.875$

Attempting to rewrite $0.875$ as a fraction multiplied by $\sqrt{3}$, we can determine values of $a$ and $b$ if needed.