A-Level Edexcel Maths Pure Solving Equations: f(x) = x³ + 3x² + 5. Find (a)

f(x) = x³ + 3x² + 5 - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 2

Question 5

f(x) = x³ + 3x² + 5. Find (a) f''(x), (b) \int_{1}^{2} f(x) \, dx.

Worked Solution & Example Answer:f(x) = x³ + 3x² + 5 - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 2

Step 1

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Answer

To find the second derivative of the function, we first need to differentiate it twice.

First Derivative $f'(x)$ :

$f'(x) = \frac{d}{dx}(x^3 + 3x^2 + 5) = 3x^2 + 6x$
Second Derivative $f''(x)$ :

$f''(x) = \frac{d}{dx}(3x^2 + 6x) = 6x + 6$

Thus, the second derivative is:

$f''(x) = 6x + 6$

Step 2

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Answer

To calculate the definite integral of $f(x)$ from 1 to 2, we perform the following steps:

Set up the integral:

$\int_{1}^{2} f(x) \, dx = \int_{1}^{2} (x^3 + 3x^2 + 5) \, dx$
Integrate the function:

$\int (x^3 + 3x^2 + 5) \, dx = \frac{x^4}{4} + x^3 + 5x + C$
Evaluate the definite integral from 1 to 2:

$\left[\frac{x^4}{4} + x^3 + 5x\right]_{1}^{2}$
- Calculate at $x = 2$ :
$\frac{2^4}{4} + 2^3 + 5(2) = \frac{16}{4} + 8 + 10 = 4 + 8 + 10 = 22$
- Calculate at $x = 1$ :
$\frac{1^4}{4} + 1^3 + 5(1) = \frac{1}{4} + 1 + 5 = \frac{1}{4} + 1 + 5 = \frac{1}{4} + \frac{4}{4} + \frac{20}{4} = \frac{25}{4}$
Final result:

$\int_{1}^{2} f(x) \, dx = 22 - \frac{25}{4} = \frac{88}{4} - \frac{25}{4} = \frac{63}{4}$

Thus, the final result is:

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