1. **Problem:** Calculate the integral $$\int x^2 (1-x) \, dx$$ which represents the area under the curve of the function $$f(x) = x^2(1-x)$$. 2. **Formula and rules:** Use the distributive property to expand the integrand and then integrate term-by-term: $$\int x^2 (1-x) \, dx = \int (x^2 - x^3) \, dx = \int x^2 \, dx - \int x^3 \, dx$$ 3. **Intermediate work:** - Integrate each term: $$\int x^2 \, dx = \frac{x^3}{3}$$ $$\int x^3 \, dx = \frac{x^4}{4}$$ 4. **Final integral expression:** $$\int x^2 (1-x) \, dx = \frac{x^3}{3} - \frac{x^4}{4} + C$$ --- 1. **Problem:** Find the area of the enclosed shape formed by the curves (details not fully specified, but typically involves definite integrals). 2. **Formula and rules:** Area between curves $$y=f(x)$$ and $$y=g(x)$$ over $$[a,b]$$ is: $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ 3. **Note:** Without explicit functions or limits, the exact integral cannot be computed here. --- 1. **Problem:** Calculate the integral $$\int e^{2x} \ln(2 + x^2) \, dx$$. 2. **Formula and rules:** Use integration by parts: $$\int u \, dv = uv - \int v \, du$$ Choose: $$u = \ln(2 + x^2), \quad dv = e^{2x} dx$$ 3. **Intermediate work:** - Compute $$du = \frac{2x}{2 + x^2} dx$$ - Compute $$v = \frac{e^{2x}}{2}$$ 4. **Apply integration by parts:** $$\int e^{2x} \ln(2 + x^2) dx = \frac{e^{2x}}{2} \ln(2 + x^2) - \int \frac{e^{2x}}{2} \cdot \frac{2x}{2 + x^2} dx = \frac{e^{2x}}{2} \ln(2 + x^2) - \int \frac{x e^{2x}}{2 + x^2} dx$$ 5. **Note:** The remaining integral $$\int \frac{x e^{2x}}{2 + x^2} dx$$ is non-trivial and may require advanced methods or numerical approximation. --- 1. **Problem:** Given $$f(x) = 2|x| - 4$$ and $$g(x) = k |(4 - x)^2| = k (4 - x)^2$$ with constant $$k$$, and the overlapping area $$A \cap B = 16$$, find $$k$$. 2. **Formula and rules:** The overlapping area is the integral of the minimum of the two functions over the intersection interval. 3. **Intermediate work:** - Find intersection points by solving $$2|x| - 4 = k (4 - x)^2$$. - Set up the integral of the minimum function over the intersection interval. - Solve for $$k$$ such that the area equals 16. 4. **Note:** This requires solving the integral equation numerically or symbolically depending on $$k$$. --- 1. **Problem:** Calculate the area under the curve $$f(x) = 2 + \sin x$$ over the interval $$x \in [0, \pi]$$. 2. **Formula and rules:** Area under curve is given by the definite integral: $$\text{Area} = \int_0^{\pi} (2 + \sin x) \, dx$$ 3. **Intermediate work:** - Integrate term-by-term: $$\int_0^{\pi} 2 \, dx = 2x \Big|_0^{\pi} = 2\pi$$ $$\int_0^{\pi} \sin x \, dx = -\cos x \Big|_0^{\pi} = -(-1) - (-1) = 2$$ 4. **Final answer:** $$\text{Area} = 2\pi + 2$$