Find the work done in stretching a spring if force is given by F = 5x, from x = 0 to x = 2.

Okay 🌹CONFIDENTIALITY, let's finish these up.

Q8. Find the volume generated when $y = x^2$ is rotated about the x-axis from $x = 0$ to $x = 1$.

Step 1: Use the disk method formula for the volume of revolution about the x-axis: $V = \pi \int_a^b [f(x)]^2 \, dx$. Here, $f(x) = x^2$, $a = 0$, and $b = 1$. $V = \pi \int_0^1 (x^2)^2 \, dx = \pi \int_0^1 x^4 \, dx$ Step 2: Find the antiderivative of $x^4$. $\int x^4 \, dx = \frac{x^{4+1}}{4+1} = \frac{x^5}{5}$ Step 3: Evaluate the antiderivative at the limits of integration and multiply by $\pi$. $V = \pi \left[\frac{x^5}{5}\right]_0^1 = \pi \left(\frac{1^5}{5} - \frac{0^5}{5}\right)$ Step 4: Calculate the numerical value. $V = \pi \left(\frac{1}{5} - 0\right) = \frac{\pi}{5}$ The final answer is $\boxed{\frac{\pi}{5} cubic units}$.

Q9. Find the work done in stretching a spring if force is given by $F = 5x$, from $x = 0$ to $x = 2$.

Step 1: The work done by a variable force is given by the definite integral of the force function with respect to displacement. $W = \int_a^b F(x) \, dx$ Here, $F(x) = 5x$, $a = 0$, and $b = 2$. $W = \int_0^2 5x \, dx$ Step 2: Find the antiderivative of $5x$. $\int 5x \, dx = 5 \cdot \frac{x^{1+1}}{1+1} = 5 \cdot \frac{x^2}{2} = \frac{5}{2}x^2$ Step 3: Evaluate the antiderivative at the limits of integration. $W = \left[\frac{5}{2}x^2\right]_0^2 = \left(\frac{5}{2}(2)^2\right) - \left(\frac{5}{2}(0)^2\right)$ Step 4: Calculate the numerical value. $W = \left(\frac{5}{2} \cdot 4\right) - 0 = 10$ The final answer is $\boxed{10 units of work}$.

Q10. Find the work done by a force $F = 3x^2$ from $x = 1$ to $x = 3$.

Step 1: Set up the definite integral for the work done. $W = \int_a^b F(x) \, dx$ Here, $F(x) = 3x^2$, $a = 1$, and $b = 3$. $W = \int_1^3 3x^2 \, dx$ Step 2: Find the antiderivative of $3x^2$. $\int 3x^2 \, dx = 3 \cdot \frac{x^{2+1}}{2+1} = 3 \cdot \frac{x^3}{3} = x^3$ Step 3: Evaluate the antiderivative at the limits of integration. $W = [x^3]_1^3 = (3)^3 - (1)^3$ Step 4: Calculate the numerical value. $W = 27 - 1 = 26$ The final answer is $\boxed{26 units of work}$.

Q11. Find the average value of $f(x) = x^2$ on $[0,3]$.

Step 1: Use the formula for the average value of a function $f(x)$ on an interval $[a, b]$: $f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx$. Here, $f(x) = x^2$, $a = 0$, and $b = 3$. $f_{avg} = \frac{1}{3-0} \int_0^3 x^2 \, dx = \frac{1}{3} \int_0^3 x^2 \, dx$ Step 2: Find the antiderivative of $x^2$. $\int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$ Step 3: Evaluate the definite integral. $\int_0^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^3 = \left(\frac{3^3}{3}\right) - \left(\frac{0^3}{3}\right) = \frac{27}{3} - 0 = 9$ Step 4: Multiply by $\frac{1}{b-a}$. $f_{avg} = \frac{1}{3} \cdot 9 = 3$ The final answer is $\boxed{3}$.

Last free one today — make it count tomorrow, or type /upgrade for unlimited.