Given the function: $$f(x) = x^3 + 3x^2 + 5.$$ Find (a) $f''(x)$, (b) $\int_{1}^{2} f(x) \, dx$. - Edexcel - A-Level Maths Pure - Question 3 - 2007 - Paper 2

To solve the integral $\int_{1}^{2} f(x) \, dx$, we substitute $f(x)$ into the integral:

$\int_{1}^{2} (x^3 + 3x^2 + 5) \, dx.$

This breaks down into three separate integrals:

$= \int_{1}^{2} x^3 \, dx + \int_{1}^{2} 3x^2 \, dx + \int_{1}^{2} 5 \, dx.$

Calculating each integral:

1. For $\int x^3 \, dx$, we have: $\frac{x^4}{4}$ evaluated from 1 to 2 gives: $\left[\frac{2^4}{4} - \frac{1^4}{4}\right] = \left[\frac{16}{4} - \frac{1}{4}\right] = \frac{15}{4}.$

2. For $\int 3x^2 \, dx$, we have: $3 \cdot \frac{x^3}{3} = x^3$ evaluated from 1 to 2 gives: $\left[2^3 - 1^3\right] = 8 - 1 = 7.$

3. For $\int 5 \, dx$, we have: $5x$ evaluated from 1 to 2 gives: $5(2) - 5(1) = 10 - 5 = 5.$

Now, summing these results together:

$= \frac{15}{4} + 7 + 5 = \frac{15}{4} + \frac{28}{4} + \frac{20}{4} = \frac{63}{4}.$

Thus, the final answer is:

$\int_{1}^{2} f(x) \, dx = \frac{63}{4}.$