✅ Manual Solution:

The area (\(A\)) of the region enclosed by the curve \(y = f(x)\), the \(x\)-axis, and the vertical lines \(x=a\) and \(x=b\) is given by the definite integral:\[A = \int_{a}^{b} f(x) dx\] \[A = \int_{0}^{4} x^{1/2} dx\] \[A = \frac{2}{3} \left[ x^{3/2} \right]_{0}^{4}\] \[A = \frac{2}{3} \left[ (4)^{3/2} - (0)^{3/2} \right]\] \[A = \frac{16}{3}\]

🤖 AI Generated Solution:

To find the area of the region, we need to integrate the function $y=\sqrt{x}$ with respect to $x$ from $x=0$ to $x=4$. 1. **Set up the integral:** The area $A$ is given by the definite integral: $$ A = \int_{0}^{4} \sqrt{x} \, dx $$ 2. **Rewrite the integrand:** $$ A = \int_{0}^{4} x^{1/2} \, dx $$ 3. **Perform the integration:** Use the power rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1}$. $$ A = \left[ \frac{x^{1/2+1}}{1/2+1} \right]_{0}^{4} = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4} = \left[ \frac{2}{3}x^{3/2} \right]_{0}^{4} $$ 4. **Evaluate the definite integral:** Substitute the upper and lower limits of integration. $$ A = \frac{2}{3}(4)^{3/2} - \frac{2}{3}(0)^{3/2} $$ 5. **Calculate the value:** $$ A = \frac{2}{3}(\sqrt{4})^3 - 0 = \frac{2}{3}(2)^3 = \frac{2}{3}(8) = \frac{16}{3} $$ The area of the region is $\frac{16}{3}$ square units. The final answer is $\boxed{\frac{16}{3}}$.

🤖 AI Generated Solution:

The function is given by $y=x|x|$. We can define this piecewise: $$ y = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases} $$ The area $A$ enclosed between the curve, the x-axis, $x=-2$ and $x=2$ is given by the integral of the absolute value of the function over the interval $[-2, 2]$: $$ A = \int_{-2}^{2} |y| \, dx = \int_{-2}^{2} |x|x|| \, dx $$ Since $|x| \ge 0$, we have $|x|x|| = |x||x| = x^2$. So, the integral becomes: $$ A = \int_{-2}^{2} x^2 \, dx $$ We can evaluate this integral: $$ A = \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{(2)^3}{3} - \frac{(-2)^3}{3} $$ $$ A = \frac{8}{3} - \frac{-8}{3} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3} $$ The final answer is $\boxed{\frac{16}{3}}$.

🤖 AI Generated Solution:

The region is bounded by the parabola $y^2=4x$, the X-axis ($y=0$), $x=0$, and $x=1$. Since the parabola is symmetric about the X-axis, the total area can be found by calculating the area of the upper half (where $y = 2\sqrt{x}$) and multiplying by 2. 1. From $y^2=4x$, we get $y = \pm \sqrt{4x} = \pm 2\sqrt{x}$. 2. The area $A$ is given by the integral of $y$ with respect to $x$ from $x=0$ to $x=1$. Due to symmetry, we calculate the area for $y=2\sqrt{x}$ and multiply by 2. $$A = 2 \int_{0}^{1} 2\sqrt{x} \, dx$$ 3. Evaluate the integral: $$A = 4 \int_{0}^{1} x^{1/2} \, dx$$ $$A = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1}$$ $$A = 4 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1}$$ 4. Apply the limits of integration: $$A = \frac{8}{3} \left( (1)^{3/2} - (0)^{3/2} \right)$$ $$A = \frac{8}{3} (1 - 0)$$ $$A = \frac{8}{3}$$ The final answer is $\boxed{\frac{8}{3}}$.

🤖 AI Generated Solution:

The given equation of the curve is $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$. This is an ellipse centered at the origin. **Step 1: Express $y$ in terms of $x$ for the upper half of the ellipse.** From the equation, we solve for $y$: $\frac{y^{2}}{4} = 1 - \frac{x^{2}}{9}$ $y^{2} = 4 \left(\frac{9-x^{2}}{9}\right)$ $y = \pm \frac{2}{3}\sqrt{9-x^{2}}$ For the region above the x-axis, we consider the positive root: $y = \frac{2}{3}\sqrt{9-x^{2}}$. **Step 2: Determine the limits of integration and set up the integral for the area.** The ellipse intersects the x-axis when $y=0$. Substituting $y=0$ into the equation gives $\frac{x^{2}}{9}=1$, so $x^{2}=9$, which means $x=\pm 3$. Due to the symmetry of the ellipse, the total area of the region bounded by the curve and the x-axis is $4$ times the area in the first quadrant (where $x$ ranges from $0$ to $3$). So, the total area $A$ is given by: $A = 4 \int_{0}^{3} y \, dx = 4 \int_{0}^{3} \frac{2}{3}\sqrt{9-x^{2}} \, dx = \frac{8}{3} \int_{0}^{3} \sqrt{9-x^{2}} \, dx$ **Step 3: Evaluate the definite integral.** We use the standard integral formula $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$. In our case, $a^2=9$, so $a=3$. $A = \frac{8}{3} \left[\frac{x}{2}\sqrt{9-x^{2}} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)\right]_{0}^{3}$ **Step 4: Calculate the area using the limits of integration.** Substitute the upper and lower limits: $A = \frac{8}{3} \left[\left(\frac{3}{2}\sqrt{9-3^{2}} + \frac{9}{2}\sin^{-1}\left(\frac{3}{3}\right)\right) - \left(\frac{0}{2}\sqrt{9-0^{2}} + \frac{9}{2}\sin^{-1}\left(\frac{0}{3}\right)\right)\right]$ $A = \frac{8}{3} \left[\left(\frac{3}{2}(0) + \frac{9}{2}\sin^{-1}(1)\right) - \left(0 + \frac{9}{2}\sin^{-1}(0)\right)\right]$ Since $\sin^{-1}(1) = \frac{\pi}{2}$ and $\sin^{-1}(0) = 0$: $A = \frac{8}{3} \left[\left(0 + \frac{9}{2} \cdot \frac{\pi}{2}\right) - (0 + 0)\right]$ $A = \frac{8}{3} \left[\frac{9\pi}{4}\right]$ $A = \frac{72\pi}{12} = 6\pi$ The area of the region bounded by the curve $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the x-axis is $6\pi$ square units. The final answer is $\boxed{6\pi}$.

🤖 AI Generated Solution:

To sketch the graph of $y = |x + 3|$: Start with the graph of $y = x + 3$, which is a straight line. The absolute value reflects any part of the graph below the $x$-axis upwards. This results in a V-shaped graph with its vertex at $(-3, 0)$. For $x \ge -3$, $y = x+3$. For $x < -3$, $y = -(x+3) = -x-3$. To find the area using integration: The area enclosed by the curve, the $x$-axis, between $x = -6$ and $x = 0$ is given by the integral: $$A = \int_{-6}^{0} |x+3| dx$$ Since the definition of $|x+3|$ changes at $x = -3$, we split the integral: $$A = \int_{-6}^{-3} -(x+3) dx + \int_{-3}^{0} (x+3) dx$$ Evaluate the first integral: $$\int_{-6}^{-3} (-x-3) dx = \left[-\frac{x^2}{2} - 3x\right]_{-6}^{-3} = \left(-\frac{9}{2} + 9\right) - \left(-\frac{36}{2} + 18\right) = \frac{9}{2} - 0 = \frac{9}{2}$$ Evaluate the second integral: $$\int_{-3}^{0} (x+3) dx = \left[\frac{x^2}{2} + 3x\right]_{-3}^{0} = (0) - \left(\frac{9}{2} - 9\right) = 0 - \left(-\frac{9}{2}\right) = \frac{9}{2}$$ The total area is the sum of these two parts: $$A = \frac{9}{2} + \frac{9}{2} = \frac{18}{2} = 9$$ The final answer is $\boxed{9}$.

🤖 AI Generated Solution:

The area bounded by the line $y=f(x)$, the x-axis, and the ordinates $x=a$ and $x=b$ is given by $A = \int_a^b |f(x)| dx$. 1. **Find the x-intercept**: Set $y=0$ to find where the line crosses the x-axis. $5x+2=0 \implies 5x=-2 \implies x = -2/5$. This x-intercept lies within the given interval $[-2, 2]$. 2. **Split the integral**: Since the function $y=5x+2$ changes sign at $x=-2/5$, we split the integral into two parts: * For $x \in [-2, -2/5]$, $5x+2 \le 0$, so $|5x+2| = -(5x+2)$. * For $x \in [-2/5, 2]$, $5x+2 \ge 0$, so $|5x+2| = 5x+2$. 3. **Set up the integral**: The total area $A$ is given by: $$A = \int_{-2}^{-2/5} -(5x+2) dx + \int_{-2/5}^{2} (5x+2) dx$$ 4. **Evaluate the integrals**: * First integral: $$\int_{-2}^{-2/5} (-5x-2) dx = \left[-\frac{5x^2}{2} - 2x\right]_{-2}^{-2/5}$$ $$= \left(-\frac{5(-2/5)^2}{2} - 2(-2/5)\right) - \left(-\frac{5(-2)^2}{2} - 2(-2)\right)$$ $$= \left(-\frac{2}{5} + \frac{4}{5}\right) - (-10 + 4) = \frac{2}{5} - (-6) = \frac{2}{5} + 6 = \frac{32}{5}$$ * Second integral: $$\int_{-2/5}^{2} (5x+2) dx = \left[\frac{5x^2}{2} + 2x\right]_{-2/5}^{2}$$ $$= \left(\frac{5(2)^2}{2} + 2(2)\right) - \left(\frac{5(-2/5)^2}{2} + 2(-2/5)\right)$$ $$= (10 + 4) - \left(\frac{2}{5} - \frac{4}{5}\right) = 14 - \left(-\frac{2}{5}\right) = 14 + \frac{2}{5} = \frac{72}{5}$$ 5. **Sum the areas**: $$A = \frac{32}{5} + \frac{72}{5} = \frac{104}{5}$$ The final answer is $\boxed{\frac{104}{5}}$.

🤖 AI Generated Solution:

The function \(y=x|x|\) can be defined piecewise: If \(x \ge 0,\) then \(|x|=x,\) so \(y = x \cdot x = x^2.\) If \(x < 0,\) then \(|x|=-x,\) so \(y = x \cdot (-x) = -x^2.\) Thus, the function is given by: $$y = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}$$ **Sketching the Graph:** For \(x \ge 0,\) the graph is the right half of the parabola \(y=x^2\) (opening upwards). For \(x < 0,\) the graph is the left half of the parabola \(y=-x^2\) (opening downwards). Both parts meet at the origin \((0,0)\). The graph is symmetric about the origin (an odd function). **Finding the Area:** The area bounded by the curve, the X-axis, and the ordinates \(x=-2\) and \(x=2\) is given by the integral \(\int_{-2}^{2} |y| dx.\) From the piecewise definition, we observe that: For \(x \ge 0,\) \(y=x^2 \ge 0,\) so \(|y|=x^2.\) For \(x < 0,\) \(y=-x^2 \le 0,\) so \(|y|=-(-x^2)=x^2.\) Therefore, for all \(x,\) \(|y|=x^2.\) The required area \(A\) is: $$A = \int_{-2}^{2} x^2 dx$$ Since \(x^2\) is an even function and the interval of integration is symmetric about the origin, we can write: $$A = 2 \int_{0}^{2} x^2 dx$$ Now, we evaluate the integral: $$A = 2 \left[ \frac{x^3}{3} \right]_{0}^{2}$$ $$A = 2 \left( \frac{2^3}{3} - \frac{0^3}{3} \right)$$ $$A = 2 \left( \frac{8}{3} - 0 \right)$$ $$A = 2 \cdot \frac{8}{3}$$ $$A = \frac{16}{3}$$ The final answer is $\boxed{\frac{16}{3}}$.