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Use calculus to find the exact value of $$\int (3x^2 + 5 + \frac{4}{x^2}) \, dx.$$ - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 2

Question 4

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Use calculus to find the exact value of $$\int (3x^2 + 5 + \frac{4}{x^2}) \, dx.$$

Worked Solution & Example Answer:Use calculus to find the exact value of $$\int (3x^2 + 5 + \frac{4}{x^2}) \, dx.$$ - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 2

Step 1

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Answer

To find the integral, we will integrate each term of the function separately:

$\int (3x^2 + 5 + \frac{4}{x^2}) \, dx = \int 3x^2 \, dx + \int 5 \, dx + \int \frac{4}{x^2} \, dx.$

For $\int 3x^2 \, dx$ , using the power rule: $= \frac{3x^{3}}{3} = x^3.$
For $\int 5 \, dx$ : $= 5x.$
For $\int \frac{4}{x^2} \, dx$ : $= 4 \int x^{-2} \, dx = 4 \cdot \left(-\frac{1}{x}\right) = -\frac{4}{x}.$

Putting it all together, we have:

$\int (3x^2 + 5 + \frac{4}{x^2}) \, dx = x^3 + 5x - \frac{4}{x} + C$

Step 2

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Answer

We use the Fundamental Theorem of Calculus to evaluate the definite integral:

Let’s assume we want to evaluate from $x = 1$ to $x = 2$ . We substitute these limits into our integrated function:

Substitute with $x = 2$ : $f(2) = (2^3) + 5(2) - \frac{4}{2} = 8 + 10 - 2 = 16.$
Substitute with $x = 1$ : $f(1) = (1^3) + 5(1) - \frac{4}{1} = 1 + 5 - 4 = 2.$

Now we find the value of the definite integral:

$\int_{1}^{2} (3x^2 + 5 + \frac{4}{x^2}) \, dx = f(2) - f(1) = 16 - 2 = 14.$

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