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

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Question 3

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

f(x) = x³ + 3x² + 5. Find (a) f''(x), (b) ∫² f(x) dx.

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

Step 1

(a) f''(x)

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Answer

To find the second derivative of the function, we start by calculating the first derivative, f'(x).

  1. Differentiate f(x).

    f′(x)=ddx(x3+3x2+5)=3x2+6xf'(x) = \frac{d}{dx}(x^3 + 3x^2 + 5) = 3x^2 + 6x

  2. Now, differentiate f'(x) to find f''(x).

    f′′(x)=ddx(3x2+6x)=6x+6f''(x) = \frac{d}{dx}(3x^2 + 6x) = 6x + 6

Thus, the second derivative is:

f′′(x)=6x+6f''(x) = 6x + 6

Step 2

(b) ∫² f(x) dx.

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Answer

To evaluate the integral of f(x) from 1 to 2, we first write the function:

f(x)=x3+3x2+5f(x) = x^3 + 3x^2 + 5

  1. Set up the integral:

    ∫12(x3+3x2+5) dx\int_{1}^{2} (x^3 + 3x^2 + 5) \, dx

  2. Integrate term by term:

    =[x44+x3+5x]12= \left[ \frac{x^4}{4} + x^3 + 5x \right]_{1}^{2}

  3. Calculate the definite integral by substituting the limits:

    =(244+23+5×2)−(144+13+5×1) = \left( \frac{2^4}{4} + 2^3 + 5 \times 2 \right) - \left( \frac{1^4}{4} + 1^3 + 5 \times 1 \right)

    =(4+8+10)−(0.25+1+5) = \left( 4 + 8 + 10 \right) - \left( 0.25 + 1 + 5 \right)

    =22−6.25=15.75 = 22 - 6.25 = 15.75

Therefore, the value of the integral is:

∫12f(x) dx=15.75\int_{1}^{2} f(x) \, dx = 15.75

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