6. (a) Find \[ \int 10x(x^2 - 2)dx \] giving each term in its simplest form. (4) (b) Figure 2 shows a sketch of part of the curve C with equation \[ y = 10x(x^2 - 2), \, x > 0 \] The curve C starts at the origin and crosses the x-axis at the point (4, 0). The area, shown shaded in Figure 2, consists of two finite regions and is bounded by the curve C, the x-axis and the line x = 9. (5) (b) Use your answer from part (a) to find the total area of the shaded regions.

A-Level Edexcel Maths Pure Basic Trigonometry: 6. (a) Find \[ \int 10x(x^2 -

6. (a) Find \[ \int 10x(x^2 - 2)dx \] giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 2

Question 8

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6. (a) Find \[ \int 10x(x^2 - 2)dx \] giving each term in its simplest form. (4) (b) Figure 2 shows a sketch of part of the curve C with equation \[... show full transcript

Worked Solution & Example Answer:6. (a) Find \[ \int 10x(x^2 - 2)dx \] giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 2

Step 1

Find $ \int 10x(x^2 - 2)dx $

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Answer

Expand the Integral:
The integrand can be expanded as follows:
[ 10x(x^2 - 2) = 10x^3 - 20x ]
Therefore, we need to compute:
[ \int (10x^3 - 20x)dx ]
Integrate Each Term:
We use the power rule of integration:
[ \int x^n dx = \frac{x^{n+1}}{n+1} + C ]
So, we integrate term by term:
[ \int 10x^3 dx = \frac{10}{4}x^4 = \frac{5}{2}x^4 ]
[ \int -20x dx = -10x^2 ] Combining these gives us:
[ \int (10x^3 - 20x)dx = \frac{5}{2}x^4 - 10x^2 + C ]
(where C is the constant of integration)
Present in Simplest Form:
The integrated result can therefore be expressed as:
[ \frac{5}{2}x^4 - 10x^2 + C ]

Step 2

Find the total area of the shaded regions

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Answer

Set Up the Area Calculation:
The total area consists of the two regions calculated separately.
The first region is from x = 0 to x = 4, and the second region is from x = 4 to x = 9.
We already found the integral which allows us to calculate the area under the curve.
Calculate Area from 0 to 4:
Using our integral result:
[\text{Area}_1 = \int_0^4 (10x^3 - 20x)dx ]
This gives
[ \left[ \frac{5}{2}x^4 - 10x^2 \right]_0^4 = \left(\frac{5}{2}(4^4) - 10(4^2)\right) - 0 = \left(\frac{5}{2}(256) - 160\right) = 640 - 160 = 480 ]
Calculate Area from 4 to 9:
Now, using the integral again from x = 4 to x = 9:
[\text{Area}_2 = \int_4^9 (10x^3 - 20x)dx ]
This simplifies to
[ \left[ \frac{5}{2}x^4 - 10x^2 \right]_4^9 = \left(\frac{5}{2}(9^4) - 10(9^2)\right) - \left(\frac{5}{2}(4^4) - 10(4^2)\right)]
Evaluating yields:
- From 9: [ \frac{5}{2}(6561) - 810 = 16402.5 - 810 = 15692.5 ]
- From 4 we previously computed to be 480
  Thus, [ \text{Area}_2 = 15692.5 - 480 = 15192.5 ]
Total Area:
Finally, summing both areas gives: [ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = 480 + 15192.5 = 15672.5 ]

6. (a) Find \[ \int 10x(x^2 - 2)dx \] giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 2

Worked Solution & Example Answer:6. (a) Find \[ \int 10x(x^2 - 2)dx \] giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 2

Find \( \int 10x(x^2 - 2)dx \)

Find the total area of the shaded regions

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