1. **Problem:** Find the area of the region bounded by the curve $y = x^2 - 2x$ and the x-axis. 2. **Formula and rules:** The area between a curve $y=f(x)$ and the x-axis from $a$ to $b$ is given by $$\text{Area} = \int_a^b |f(x)| \, dx.$$ Since the region is below the x-axis, we take the absolute value of the function. 3. **Find the points of intersection with the x-axis:** Solve $x^2 - 2x = 0$. $$x(x - 2) = 0 \implies x = 0 \text{ or } x = 2.$$ 4. **Set up the integral:** The function is negative between $0$ and $2$, so the area is $$\int_0^2 |x^2 - 2x| \, dx = \int_0^2 (2x - x^2) \, dx.$$ 5. **Calculate the integral:** $$\int_0^2 (2x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = \left(4 - \frac{8}{3}\right) - 0 = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}.$$ 6. **Answer:** The area of the region is $\boxed{\frac{4}{3}}$. --- 1. **Problem:** Find the area of the region bounded by the lines $y = x$, $y = -x$, and $x = 1$. 2. **Description:** The region is a triangle bounded by the lines $y = x$, $y = -x$, and the vertical line $x=1$. 3. **Find the points of intersection:** - $y = x$ and $y = -x$ intersect at $(0,0)$. - The vertical line $x=1$ intersects $y=x$ at $(1,1)$ and $y=-x$ at $(1,-1)$. 4. **Calculate the area:** The base is the segment from $(1,-1)$ to $(1,1)$ with length $2$. The height is the distance from $(0,0)$ to $x=1$, which is $1$. Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$. 5. **Answer:** The area of the region is $\boxed{1}$. --- 1. **Problem:** Find the area of the region bounded by the curves $y = x^2$, $y = -x^2$, and the line $x = 1$. 2. **Description:** The region is symmetric about the x-axis between $x=0$ and $x=1$. 3. **Set up the integral:** The area is twice the area above the x-axis: $$2 \int_0^1 x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_0^1 = 2 \times \frac{1}{3} = \frac{2}{3}.$$ 4. **Answer:** The area of the region is $\boxed{\frac{2}{3}}$. --- 1. **Problem:** Find the area of the region bounded by the x-axis and the curve $y = -x^2 + 4x - 3$. 2. **Find the roots:** Solve $-x^2 + 4x - 3 = 0$. Multiply both sides by $-1$: $$\cancel{-}x^2 + 4x - 3 = 0 \implies \cancel{-}(-x^2 + 4x - 3) = 0 \implies x^2 - 4x + 3 = 0.$$ 3. **Factor:** $$x^2 - 4x + 3 = (x - 3)(x - 1) = 0 \implies x = 1, 3.$$ 4. **Set up the integral:** The curve is above the x-axis between $1$ and $3$, so area is $$\int_1^3 (-x^2 + 4x - 3) \, dx.$$ 5. **Calculate the integral:** $$\int_1^3 (-x^2 + 4x - 3) \, dx = \left[-\frac{x^3}{3} + 2x^2 - 3x \right]_1^3 = \left(-\frac{27}{3} + 18 - 9\right) - \left(-\frac{1}{3} + 2 - 3\right) = (-9 + 18 - 9) - (-\frac{1}{3} -1) = 0 - (-\frac{4}{3}) = \frac{4}{3}.$$ 6. **Answer:** The area of the region is $\boxed{\frac{4}{3}}$. --- 1. **Problem:** Find the area of the region bounded by the curves $y = x^2 + 1$ and $y = 2x - 2$. 2. **Find points of intersection:** Solve $x^2 + 1 = 2x - 2$. $$x^2 + 1 = 2x - 2 \implies x^2 - 2x + 3 = 0.$$ 3. **Calculate discriminant:** $$\Delta = (-2)^2 - 4 \times 1 \times 3 = 4 - 12 = -8 < 0,$$ so no real intersection points; the curves do not intersect. 4. **Check relative positions:** For $x=0$, $y = 1$ for parabola and $y = -2$ for line, so parabola is above line. 5. **Since no intersection, no bounded region exists between these curves.** 6. **Answer:** No bounded region exists; area is $\boxed{0}$. --- 1. **Problem:** Find the area of the triangle bounded by the line $y = 4 - 3x$ and the coordinate axes. 2. **Find intercepts:** - $y$-intercept at $x=0$: $y = 4$. - $x$-intercept at $y=0$: $0 = 4 - 3x \implies x = \frac{4}{3}$. 3. **Calculate area:** Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{4}{3} \times 4 = \frac{8}{3}$. 4. **Answer:** The area is $\boxed{\frac{8}{3}}$. --- 1. **Problem:** Find the area of the triangle with vertices $(-4,0)$, $(2,0)$, and $(2,6)$. 2. **Calculate base and height:** - Base length between $(-4,0)$ and $(2,0)$ is $2 - (-4) = 6$. - Height is the vertical distance from $(2,0)$ to $(2,6)$, which is $6$. 3. **Calculate area:** Area = $\frac{1}{2} \times 6 \times 6 = 18$. 4. **Answer:** The area is $\boxed{18}$. --- 1. **Problem:** Find the area of the rectangle with vertices $(1,0)$, $(-2,0)$, $(-2,5)$, and $(1,5)$. 2. **Calculate length and width:** - Length along x-axis: $1 - (-2) = 3$. - Width along y-axis: $5 - 0 = 5$. 3. **Calculate area:** Area = $3 \times 5 = 15$. 4. **Answer:** The area is $\boxed{15}$. ---