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The curve C has equation $y = x^2 - 5x + 4$ - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 4

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The curve C has equation $y = x^2 - 5x + 4$. It cuts the x-axis at the points L and M as shown in Figure 2. (a) Find the coordinates of the point L and the point M.... show full transcript

Worked Solution & Example Answer:The curve C has equation $y = x^2 - 5x + 4$ - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 4

Step 1

Find the coordinates of the point L and the point M.

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Answer

To find the points where the curve intersects the x-axis, we set y=0y = 0:

0=x25x+40 = x^2 - 5x + 4

Factoring gives:

(x1)(x4)=0(x - 1)(x - 4) = 0

Thus, the points where the curve intersects the x-axis are x=1x = 1 and x=4x = 4. Therefore, the coordinates are:

  • Point L: (1,0)(1, 0)
  • Point M: (4,0)(4, 0)

Step 2

Show that the point N(5, 4) lies on C.

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Answer

To show that the point N(5, 4) lies on curve C, we substitute x=5x = 5 into the equation:

y=525(5)+4=2525+4=4y = 5^2 - 5(5) + 4 = 25 - 25 + 4 = 4

Since y=4y = 4 when x=5x = 5, the point N(5, 4) indeed lies on the curve.

Step 3

Find $\\int (x^2 - 5x + 4) \,dx$.

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Answer

We can find the integral:

int(x25x+4)dx=x335x22+4x+C\\int (x^2 - 5x + 4) \,dx = \frac{x^3}{3} - \frac{5x^2}{2} + 4x + C

where C is the constant of integration.

Step 4

Use your answer to part (c) to find the exact value of the area of R.

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Answer

To find the area of region R, we need to calculate the definite integral of the curve between points L and M. The limits are from x=1x = 1 to x=4x = 4:

Area=14(x25x+4)dx\text{Area} = \int_1^4 (x^2 - 5x + 4) \,dx\,

Substituting the limits using the indefinite integral found in part (c):

=[x335x22+4x]14= \left[\frac{x^3}{3} - \frac{5x^2}{2} + 4x\right]_1^4

Calculating this gives:

=(4335(42)2+4(4))(1335(12)2+4(1))= \left(\frac{4^3}{3} - \frac{5(4^2)}{2} + 4(4)\right) - \left(\frac{1^3}{3} - \frac{5(1^2)}{2} + 4(1)\right)

Evaluating both parts results in the exact value of the area of R.

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