1. **Evaluate the integral** $$\int_0^1 \int_1^2 x(x + y) \, dy \, dx$$. 2. **Determine the integral** $$\int_{-1}^2 (x^3 - 2x) \, dx$$. 3. **Determine using Bernoulli's formula** the integral $$\int x^4 \sin x \, dx$$. 4. **Evaluate the integral** $$\int \frac{dx}{e^x - 1}$$. 5. **Evaluate the integral** $$\int_{-2}^2 |x + 1| \, dx$$. --- ### Step 1: Evaluate $$\int_0^1 \int_1^2 x(x + y) \, dy \, dx$$ 1. The problem asks to evaluate the double integral over $$x$$ from 0 to 1 and $$y$$ from 1 to 2 of the function $$x(x + y)$$. 2. The formula for double integrals is $$\int_a^b \int_c^d f(x,y) \, dy \, dx$$. 3. First, integrate with respect to $$y$$: $$\int_1^2 x(x + y) \, dy = x \int_1^2 (x + y) \, dy = x \left[ xy + \frac{y^2}{2} \right]_1^2$$ 4. Evaluate the inner integral: $$= x \left[ x(2) + \frac{2^2}{2} - \left(x(1) + \frac{1^2}{2}\right) \right] = x \left[ 2x + 2 - x - \frac{1}{2} \right] = x \left( x + \frac{3}{2} \right) = x^2 + \frac{3}{2} x$$ 5. Now integrate with respect to $$x$$ from 0 to 1: $$\int_0^1 \left(x^2 + \frac{3}{2} x\right) \, dx = \left[ \frac{x^3}{3} + \frac{3}{4} x^2 \right]_0^1 = \frac{1}{3} + \frac{3}{4} = \frac{4}{12} + \frac{9}{12} = \frac{13}{12}$$ **Answer:** $$\frac{13}{12}$$ --- ### Step 2: Determine $$\int_{-1}^2 (x^3 - 2x) \, dx$$ 1. The problem asks to evaluate the definite integral of $$x^3 - 2x$$ from $$-1$$ to $$2$$. 2. Use the power rule for integration: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ 3. Integrate: $$\int (x^3 - 2x) \, dx = \frac{x^4}{4} - x^2 + C$$ 4. Evaluate from $$-1$$ to $$2$$: $$\left[ \frac{x^4}{4} - x^2 \right]_{-1}^2 = \left( \frac{2^4}{4} - 2^2 \right) - \left( \frac{(-1)^4}{4} - (-1)^2 \right) = \left( \frac{16}{4} - 4 \right) - \left( \frac{1}{4} - 1 \right) = (4 - 4) - \left( \frac{1}{4} - 1 \right) = 0 - \left( -\frac{3}{4} \right) = \frac{3}{4}$$ **Answer:** $$\frac{3}{4}$$ --- ### Step 3: Determine using Bernoulli's formula $$\int x^4 \sin x \, dx$$ 1. Bernoulli's formula for repeated integration by parts is: $$\int x^n f^{(n)}(x) \, dx = x^n f^{(n-1)}(x) - n \int x^{n-1} f^{(n-1)}(x) \, dx$$ 2. Here, we apply integration by parts repeatedly with: - $$u = x^4$$, so $$du = 4x^3 dx$$ - $$dv = \sin x dx$$, so $$v = -\cos x$$ 3. First integration by parts: $$\int x^4 \sin x \, dx = -x^4 \cos x + 4 \int x^3 \cos x \, dx$$ 4. Repeat integration by parts on $$\int x^3 \cos x \, dx$$, and continue similarly until the power of $$x$$ reduces to zero. 5. The full expansion is lengthy; the final answer is: $$\int x^4 \sin x \, dx = -x^4 \cos x + 4x^3 \sin x - 12x^2 \cos x - 24x \sin x + 24 \cos x + C$$ --- ### Step 4: Evaluate $$\int \frac{dx}{e^x - 1}$$ 1. This integral is known as the integral of the Bose-Einstein distribution function. 2. It does not have an elementary antiderivative expressible in elementary functions. 3. It can be expressed in terms of the polylogarithm function or special functions. 4. Therefore, the integral is: $$\int \frac{dx}{e^x - 1} = \ln|1 - e^{-x}| + C$$ (Using substitution and properties of logarithms.) --- ### Step 5: Evaluate $$\int_{-2}^2 |x + 1| \, dx$$ 1. The absolute value function splits the integral at the point where $$x + 1 = 0$$, i.e., at $$x = -1$$. 2. Split the integral: $$\int_{-2}^2 |x + 1| \, dx = \int_{-2}^{-1} -(x + 1) \, dx + \int_{-1}^2 (x + 1) \, dx$$ 3. Evaluate the first integral: $$\int_{-2}^{-1} -(x + 1) \, dx = \int_{-2}^{-1} (-x - 1) \, dx = \left[ -\frac{x^2}{2} - x \right]_{-2}^{-1} = \left(-\frac{1}{2} + 1\right) - \left(-2 - 2\right) = \frac{1}{2} + 4 = \frac{9}{2}$$ 4. Evaluate the second integral: $$\int_{-1}^2 (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{-1}^2 = \left( 2 + 2 \right) - \left( \frac{1}{2} - 1 \right) = 4 - \left(-\frac{1}{2}\right) = \frac{9}{2}$$ 5. Sum both parts: $$\frac{9}{2} + \frac{9}{2} = 9$$ **Answer:** $$9$$