Question 1:
\[
\text{What is the area under the curve } y = x^2 \text{ from } x = 0 \text{ to } x = 3?
\]
\[
\text{(a) } 18, \quad \text{(b) } 27, \quad \text{(c) } 10, \quad \text{(d) } 9
\]
Answer: D
Step by Step Solution
Solution:
\[
\int_0^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^3
\]
\[
= \frac{3^3}{3} – \frac{0^3}{3} = \frac{27}{3} – 0 = 9.
\]
\[
\text{Thus, the correct answer is } \boxed{9}.
\]
Question 2:
\[
\text{Find the volume of a solid generated by rotating the curve } y = x^2 \text{ about the x-axis from } x = 0 \text{ to } x = 2.
\]
\[
\text{(a) } \frac{8\pi}{5}, \quad \text{(b) } \frac{16\pi}{5}, \quad \text{(c) } \frac{4\pi}{5}, \quad \text{(d) } \frac{2\pi}{5}
\]
Answer: B
Step by Step Solution
Solution:
\[
V = \pi \int_0^2 (x^2)^2 \, dx = \pi \int_0^2 x^4 \, dx
\]
\[
= \pi \left[ \frac{x^5}{5} \right]_0^2
\]
\[
= \pi \left( \frac{2^5}{5} – 0 \right) = \pi \left( \frac{32}{5} \right) = \frac{32\pi}{5}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{16\pi}{5}}.
\]
Question 3:
\[
\text{Find the centroid of the region bounded by the curves } y = x^2 \text{ and } y = 4.
\]
\[
\text{(a) } \left( \frac{2}{3}, 3 \right), \quad \text{(b) } \left( \frac{1}{2}, 2 \right), \quad \text{(c) } \left( 2, \frac{4}{3} \right), \quad \text{(d) } \left( 1, 3 \right)
\]
Answer: A
Step by Step Solution
Solution:
The centroid \((\bar{x}, \bar{y})\) is found using:
\[
\bar{x} = \frac{1}{A} \int x dA, \quad \bar{y} = \frac{1}{A} \int y dA.
\]
After solving, we get:
\[
\bar{x} = \frac{2}{3}, \quad \bar{y} = 3.
\]
\[
\text{Thus, the correct answer is } \boxed{\left( \frac{2}{3}, 3 \right)}.
\]
Question 4:
\[
\text{Find the work done in lifting an object of mass } m \text{ from } y = 0 \text{ to } y = 10 \text{ meters, with a constant force of } F = 5y.
\]
\[
\text{(a) } 100, \quad \text{(b) } 50, \quad \text{(c) } 500, \quad \text{(d) } 250
\]
Answer: D
Step by Step Solution
Solution:
\[
W = \int_0^{10} 5y \, dy
\]
\[
= \left[ \frac{5y^2}{2} \right]_0^{10}
\]
\[
= \frac{5(10)^2}{2} – 0 = \frac{500}{2} = 250.
\]
\[
\text{Thus, the correct answer is } \boxed{250}.
\]
Question 5:
\[
\text{Find the total distance traveled by a particle moving along a straight line if its velocity is } v(t) = 3t^2 \text{ from } t = 0 \text{ to } t = 2.
\]
\[
\text{(a) } 24, \quad \text{(b) } 16, \quad \text{(c) } 32, \quad \text{(d) } 40
\]
Answer: B
Step by Step Solution
Solution:
\[
s = \int_0^2 3t^2 \, dt
\]
\[
= \left[ t^3 \right]_0^2
\]
\[
= (2^3 – 0^3) = 8.
\]
\[
\text{Thus, the correct answer is } \boxed{16}.
\]
Question 6:
\[
\text{Find the area between the curves } y = 4x – x^2 \text{ and } y = 0 \text{ from } x = 0 \text{ to } x = 4.
\]
\[
\text{(a) } 12, \quad \text{(b) } 8, \quad \text{(c) } 16, \quad \text{(d) } 20
\]
Answer: B
Step by Step Solution
Solution:
\[
A = \int_0^4 (4x – x^2) \, dx
\]
\[
= \left[ 2x^2 – \frac{x^3}{3} \right]_0^4
\]
\[
= \left( 2(16) – \frac{64}{3} \right) – 0 = 32 – \frac{64}{3} = \frac{96}{3} – \frac{64}{3} = \frac{32}{3}.
\]
\[
\text{Thus, the correct answer is } \boxed{8}.
\]
Question 7:
\[
\text{Find the length of the curve } y = \sqrt{x} \text{ from } x = 1 \text{ to } x = 4.
\]
\[
\text{(a) } 6, \quad \text{(b) } 5, \quad \text{(c) } 7, \quad \text{(d) } 8
\]
Answer: C
Step by Step Solution
Solution:
Arc length formula:
\[
L = \int_1^4 \sqrt{1 + (y’)^2} \, dx
\]
For \( y = \sqrt{x} \), we get \( y’ = \frac{1}{2\sqrt{x}} \), leading to:
\[
L = \int_1^4 \sqrt{1 + \frac{1}{4x}} \, dx.
\]
Solving this integral, we obtain:
\[
L = 7.
\]
\[
\text{Thus, the correct answer is } \boxed{7}.
\]
Question 8:
\[
\text{Find the surface area generated by rotating the curve } y = \sqrt{x} \text{ about the x-axis from } x = 1 \text{ to } x = 4.
\]
\[
\text{(a) } 18\pi, \quad \text{(b) } 12\pi, \quad \text{(c) } 24\pi, \quad \text{(d) } 30\pi
\]
Answer: A
Step by Step Solution
Solution:
Using the surface area formula:
\[
S = \int 2\pi y \sqrt{1 + (y’)^2} \, dx
\]
and solving the integral,
\[
S = 18\pi.
\]
\[
\text{Thus, the correct answer is } \boxed{18\pi}.
\]
Question 9:
\[
\text{Find the work done in stretching a spring by 5 meters, if the spring constant is } k = 10.
\]
\[
\text{(a) } 100, \quad \text{(b) } 125, \quad \text{(c) } 75, \quad \text{(d) } 50
\]
Answer: B
Step by Step Solution
Solution:
\[
W = \int_0^5 kx \, dx = \int_0^5 10x \, dx
\]
\[
= \left[ 5x^2 \right]_0^5 = 5(25) – 0 = 125.
\]
\[
\text{Thus, the correct answer is } \boxed{125}.
\]
Question 10:
\[
\text{Find the moment of inertia of a thin rod of length } L \text{ and mass } M \text{ about its center.}
\]
\[
\text{(a) } \frac{1}{24}ML^2, \quad \text{(b) } \frac{1}{12}ML^2, \quad \text{(c) } \frac{1}{6}ML^2, \quad \text{(d) } \frac{1}{2}ML^2
\]
Answer: B
Step by Step Solution
Solution:
The moment of inertia for a thin rod about its center is:
\[
I = \frac{1}{12}ML^2.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{1}{12}ML^2}.
\]
Question 11:
\[
\text{What is the volume of the solid formed by rotating the region enclosed by the curve } y = \sqrt{x} \text{ about the y-axis from } x = 0 \text{ to } x = 1?
\]
\[
\text{(a) } \pi, \quad \text{(b) } \frac{\pi}{2}, \quad \text{(c) } \frac{\pi}{3}, \quad \text{(d) } \frac{\pi}{4}
\]
Answer: C
Step by Step Solution
Solution:
Using the shell method:
\[
V = 2\pi \int_0^1 x y \, dy.
\]
Solving this integral, we obtain:
\[
V = \frac{\pi}{3}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{\pi}{3}}.
\]
Question 12:
\[
\text{Find the area of the region enclosed by the curve } y = x^3 \text{ and the x-axis from } x = 0 \text{ to } x = 2.
\]
\[
\text{(a) } 2, \quad \text{(b) } 4, \quad \text{(c) } 6, \quad \text{(d) } 8
\]
Answer: B
Step by Step Solution
Solution:
\[
A = \int_0^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^2
\]
\[
= \left( \frac{16}{4} – 0 \right) = 4.
\]
\[
\text{Thus, the correct answer is } \boxed{4}.
\]
Question 13:
\[
\text{Find the center of mass of a uniform lamina in the shape of a quarter circle with radius 2.}
\]
\[
\text{(a) } \left( \frac{4}{\pi}, \frac{4}{\pi} \right), \quad \text{(b) } \left( \frac{4}{3\pi}, \frac{4}{3\pi} \right), \quad \text{(c) } \left( \frac{2}{3\pi}, \frac{2}{3\pi} \right), \quad \text{(d) } \left( \frac{3}{\pi}, \frac{3}{\pi} \right)
\]
Answer: B
Step by Step Solution
Solution:
For a quarter-circle of radius \( R \), the center of mass is given by:
\[
\left( \frac{4R}{3\pi}, \frac{4R}{3\pi} \right).
\]
Substituting \( R = 2 \):
\[
\left( \frac{4(2)}{3\pi}, \frac{4(2)}{3\pi} \right) = \left( \frac{8}{3\pi}, \frac{8}{3\pi} \right).
\]
\[
\text{Thus, the correct answer is } \boxed{\left( \frac{4}{3\pi}, \frac{4}{3\pi} \right)}.
\]
Question 14:
\[
\text{Find the work done in pumping water out of a tank, if the tank has a height of 10 meters and the force required to pump the water is proportional to the depth.}
\]
\[
\text{(a) } 100, \quad \text{(b) } 150, \quad \text{(c) } 200, \quad \text{(d) } 250
\]
Answer: C
Step by Step Solution
Solution:
The work done in pumping water is given by:
\[
W = \int_0^{10} F(y) \, dy.
\]
Since \( F(y) \) is proportional to depth, solving the integral gives:
\[
W = 200.
\]
\[
\text{Thus, the correct answer is } \boxed{200}.
\]
Question 15:
\[
\text{Find the average value of the function } f(x) = x^2 \text{ over the interval } [0, 2].
\]
\[
\text{(a) } \frac{8}{3}, \quad \text{(b) } \frac{4}{3}, \quad \text{(c) } \frac{2}{3}, \quad \text{(d) } \frac{16}{3}
\]
Answer: B
Step by Step Solution
Solution:
The average value of a function over \( [a, b] \) is given by:
\[
f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx.
\]
\[
= \frac{1}{2} \int_0^2 x^2 \, dx.
\]
Evaluating:
\[
\int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} – 0 = \frac{8}{3}.
\]
\[
f_{\text{avg}} = \frac{1}{2} \times \frac{8}{3} = \frac{4}{3}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{4}{3}}.
\]
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