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secondary 4 | A Maths
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Date Posted: 4 months ago
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done
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clear
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integral to find area under graph
Date Posted:
4 months ago
the ans is 28/15 for part 2 in case it is not clear in the pic
1 Find the derivative of the curve equation to get the equation of the tangent line.
The derivative of y = x^2 + 4x - 5 is y' = 2x + 4.
2 Set the derivative equal to the given gradient of 10 and solve for x.
2x + 4 = 10
2x = 6
x = 3
3 Substitute the value of x into the original curve equation to find the y-coordinate of P.
y = (3)^2 + 4(3) - 5
y = 9 + 12 - 5
y = 16
4 Find the x-coordinate of R by solving the curve equation for x when y = 0.
0 = x^2 + 4x - 5
(x + 5)(x - 1) = 0
x = -5 or x = 1
Since x > 0, x = 1.
5 Find the x-coordinate of Q by substituting the y-coordinate of P into the equation of the tangent line.
y - 16 = 10(x - 3)
0 - 16 = 10(x - 3)
-16 = 10(x - 3)
-16 = 10x - 30
10x = 14
x = 7/5
6 Calculate the area of the shaded region PQR.
Area of PQR = Area of PRS - Area of triangle PQS
Area of PRS = ∫(1 to 3) (x^2 + 4x - 5) dx
Area of PRS = (1/3)x^3 + 2x^2 - 5x | (1 to 3)
Area of PRS = [(1/3)(3)^3 + 2(3)^2 - 5(3)] - [(1/3)(1)^3 + 2(1)^2 - 5(1)]
Area of PRS = (27/3 + 18 - 15) - (1/3
The derivative of y = x^2 + 4x - 5 is y' = 2x + 4.
2 Set the derivative equal to the given gradient of 10 and solve for x.
2x + 4 = 10
2x = 6
x = 3
3 Substitute the value of x into the original curve equation to find the y-coordinate of P.
y = (3)^2 + 4(3) - 5
y = 9 + 12 - 5
y = 16
4 Find the x-coordinate of R by solving the curve equation for x when y = 0.
0 = x^2 + 4x - 5
(x + 5)(x - 1) = 0
x = -5 or x = 1
Since x > 0, x = 1.
5 Find the x-coordinate of Q by substituting the y-coordinate of P into the equation of the tangent line.
y - 16 = 10(x - 3)
0 - 16 = 10(x - 3)
-16 = 10(x - 3)
-16 = 10x - 30
10x = 14
x = 7/5
6 Calculate the area of the shaded region PQR.
Area of PQR = Area of PRS - Area of triangle PQS
Area of PRS = ∫(1 to 3) (x^2 + 4x - 5) dx
Area of PRS = (1/3)x^3 + 2x^2 - 5x | (1 to 3)
Area of PRS = [(1/3)(3)^3 + 2(3)^2 - 5(3)] - [(1/3)(1)^3 + 2(1)^2 - 5(1)]
Area of PRS = (27/3 + 18 - 15) - (1/3
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