I have three questions which I apparently have 'wrong' according to the book, but I've done lots of other questions which I've gotten right using the same method and I have triple checked these answers for these three questions so I was wondering if someone could tell me if I actually do have these answers wrong or if its the book?
Find the areas enclosed by the x-axis and these curves and straight lines
a) y= (1/8)x^3 + 2x, x=2, x=4
for this I got 132 sq units, but the book says 19.5
b) y=6-(1/(sqrt x)), x=16, x=25
for this I got 53 sq units and in the book it says 52
c) y= (3x-4)^2, x=1, x=3
for this I got 13 sq units but the book says its 14 sq units
Thanks in advance!
a) The integral of the function from 2 to 4 is the difference in values of
x^4/32 + x^2 at the two points. That would be 8 + 16 - 0.5 - 4 = 19.5
b) The indefinite integral is
6x - 2 x^1/2
The difference in that function at the two limits is
150 - 10 - 96 + 8 = 52
I suggest you check your math again, or the way you are doing the integration.
To verify if your answers or the book's answers are correct, let's go through the process of finding the areas enclosed by the given curves and straight lines.
a) To find the area enclosed by the curve y = (1/8)x^3 + 2x, x = 2, and x = 4, we can use the definite integral:
A = ∫[2,4] ((1/8)x^3 + 2x) dx
Evaluating this integral, we get:
A = [((1/32)x^4)/4 + x^2] evaluated from x = 2 to x = 4
= (1/128)(4^4)/4 + 4^2 - (1/128)(2^4)/4 - 2^2
= (1/32)(64)/4 + 16 - (1/32)(16)/4 - 4
= (2/4) + 16 - (1/4) - 4
= 1/2 + 16 - 1/4 - 4
= 1/2 + 16 - 1 - 4
= 19.5 sq units
Therefore, your answer of 132 sq units is incorrect, and the book's answer of 19.5 sq units is correct.
b) To find the area enclosed by the curve y = 6 - (1/√x), x = 16, and x = 25, we can again use the definite integral:
A = ∫[16,25] (6 - (1/√x)) dx
Evaluating this integral, we get:
A = [6x - 2(√x)] evaluated from x = 16 to x = 25
= (6*25 - 2(√25)) - (6*16 - 2(√16))
= (150 - 2(5)) - (96 - 2(4))
= 150 - 10 - 96 + 8
= 52 sq units
Your answer of 53 sq units is correct, and the book's answer of 52 sq units is also correct.
c) To find the area enclosed by the curve y = (3x - 4)^2, x = 1, and x = 3, we can once again use the definite integral:
A = ∫[1,3] ((3x - 4)^2) dx
Evaluating this integral, we get:
A = [(3/3)(3x - 4)^3] evaluated from x = 1 to x = 3
= [(3/3)((3(3) - 4)^3) - (3/3)((3(1) - 4)^3)]
= [(3/3)(5^3) - (3/3)(-1^3)]
= [(3/3)(125) - (3/3)(-1)]
= 125 + 1
= 126 sq units
Your answer of 13 sq units is incorrect, and the book's answer of 14 sq units is also incorrect.
To summarize:
a) The correct area is 19.5 sq units.
b) Both your answer (53 sq units) and the book's answer (52 sq units) are correct.
c) The correct area is 126 sq units.
It seems like you made a mistake in calculating the area for question c. Make sure to double-check your calculations.