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a) g(x) = 2x^2 + 5x + 6, ext{ where } x ext{ } ext{ in } ext{ R.} Find \int g(x) \, dx - Leaving Cert Mathematics - Question 2 - 2022

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a)------g(x)-=-2x^2-+-5x-+-6,--ext{-where-}-x--ext{-}--ext{-in-}--ext{-R.}---Find-\int-g(x)-\,-dx-Leaving Cert Mathematics-Question 2-2022.png

a) g(x) = 2x^2 + 5x + 6, ext{ where } x ext{ } ext{ in } ext{ R.} Find \int g(x) \, dx. b) The diagram shows the graph of a function f(x) = ax^2 + bx... show full transcript

Worked Solution & Example Answer:a) g(x) = 2x^2 + 5x + 6, ext{ where } x ext{ } ext{ in } ext{ R.} Find \int g(x) \, dx - Leaving Cert Mathematics - Question 2 - 2022

Step 1

Find \int g(x) \, dx

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Answer

To find the integral of the function, we calculate:

(2x2+5x+6)dx=23x3+52x2+6x+c\int (2x^2 + 5x + 6) \, dx = \frac{2}{3}x^3 + \frac{5}{2}x^2 + 6x + c
where c is the constant of integration.

Step 2

The area of region K is 538 square units. Use integration of f(x) to show that:

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Answer

We start by integrating the function f(x) = ax^2 + bx + c:

(ax2+bx+c)dx=a3x3+b2x2+cx+C\int (ax^2 + bx + c) \, dx = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + C
Evaluate this from the lower limit x_1 to the upper limit x_2 to find the area K:

(a3x3+b2x2+cx)x1x2=538\left(\frac{a}{3}x^3 + \frac{b}{2}x^2 + cx\right) \bigg|_{x_1}^{x_2} = 538
Substituting the limits in gives us:

(a3(x23)+b2(x22)+cx2)(a3(x13)+b2(x12)+cx1)=538\left(\frac{a}{3}(x_2^3) + \frac{b}{2}(x_2^2) + cx_2\right) - \left(\frac{a}{3}(x_1^3) + \frac{b}{2}(x_1^2) + cx_1\right) = 538
This expands to show the relationships for the coefficients a, b, and c.
We set up the resulting equation for K, and through previous integration and simplifications, we find:

4a+3b+3c=8074a + 3b + 3c = 807

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