Let's solve each double integral step by step. --- ### (a) \(\int_1^4 \left( \int_1^2 \left( \frac{2x}{y} - \frac{1}{\sqrt{y}} \right) dy \right) dx\) #### **Step 1: Integrate with respect to \(y\) first:** \[ \int_1^2 \left( \frac{2x}{y} - \frac{1}{\sqrt{y}} \right) dy \] Integrate term by term: - \(\int \frac{2x}{y} dy = 2x \int \frac{1}{y} dy = 2x \ln y\) - \(\int \frac{1}{\sqrt{y}} dy = \int y^{-1/2} dy = 2y^{1/2}\) So, \[ \int_1^2 \left( \frac{2x}{y} - \frac{1}{\sqrt{y}} \right) dy = \left[2x \ln y - 2y^{1/2}\right]_1^2 \] Plug in \(y=2\) and \(y=1\): - At \(y=2\): \(2x \ln 2 - 2(\sqrt{2})\) - At \(y=1\): \(2x \ln 1 - 2(1)\) \(= 0 - 2\) So the difference is: \[ \left(2x \ln 2 - 2\sqrt{2}\right) - \left(0 - 2\right) = 2x \ln 2 - 2\sqrt{2} + 2 \] --- #### **Step 2: Integrate with respect to \(x\):** \[ \int_1^4 \left(2x \ln 2 - 2\sqrt{2} + 2\right) dx \] Integrate term by term: - \(\int_1^4 2x \ln 2 dx = 2 \ln 2 \int_1^4 x dx = 2 \ln 2 \left[\frac{x^2}{2}\right]_1^4 = 2 \ln 2 \left(\frac{16}{2} - \frac{1}{2}\right) = 2 \ln 2 \left(8 - 0.5\right) = 2 \ln 2 \cdot 7.5 = 15 \ln 2\) - \(\int_1^4 -2\sqrt{2} dx = -2\sqrt{2} (4-1) = -2\sqrt{2} \cdot 3 = -6\sqrt{2}\) - \(\int_1^4 2 dx = 2(4-1) = 6\) --- #### **Step 3: Final answer for (a):** \[ \boxed{15 \ln 2 - 6\sqrt{2} + 6} \] --- ### (b) \(\int_3^5 \int_1^y \frac{x \ln y}{y} dx dy\) #### **Step 1: Integrate with respect to \(x\):** \[ \int_1^y \frac{x \ln y}{y} dx \] \(\frac{\ln y}{y}\) is a constant with respect to \(x\): \[ = \frac{\ln y}{y} \int_1^y x dx = \frac{\ln y}{y} \left[\frac{x^2}{2}\right]_1^y = \frac{\ln y}{y} \left(\frac{y^2}{2} - \frac{1}{2}\right) = \frac{\ln y}{y} \cdot \frac{y^2 - 1}{2} \] \[ = \frac{(y^2 - 1)\ln y}{2y} \] --- #### **Step 2: Integrate with respect to \(y\):** \[ \int_3^5 \frac{(y^2 - 1)\ln y}{2y} dy = \frac{1}{2} \int_3^5 \frac{(y^2 - 1)\ln y}{y} dy \] Break up \(y^2 - 1\): \[ = \frac{1}{2} \int_3^5 (y \ln y - \frac{\ln y}{y}) dy \] Integrate: - \(\int y \ln y dy\) (Integration by parts): Let \(u = \ln y, dv = y dy\), \(du = \frac{1}{y} dy, v = \frac{y^2}{2}\): \[ \int y \ln y dy = \frac{y^2}{2} \ln y - \int \frac{y^2}{2} \cdot \frac{1}{y} dy = \frac{y^2}{2} \ln y - \frac{1}{2} \int y dy = \frac{y^2}{2} \ln y - \frac{1}{4} y^2 \] - \(\int \frac{\ln y}{y} dy = \frac{1}{2} (\ln y)^2\) So, plug in: \[ \frac{1}{2} \left[ \left( \frac{y^2}{2}\ln y - \frac{1}{4} y^2 \right) - \frac{1}{2} (\ln y)^2 \right]_3^5 \] --- #### **Step 3: Plug in the limits:** Let’s write this clearly: \[ \frac{1}{2} \left\{ \left[ \frac{y^2}{2} \ln y - \frac{1}{4}y^2 - \frac{1}{2} (\ln y)^2 \right]_{y=3}^{y=5} \right\} \] So, \[ = \frac{1}{2} \left( \left[ \frac{25}{2} \ln 5 - \frac{25}{4} - \frac{1}{2} (\ln 5)^2 \right] - \left[ \frac{9}{2} \ln 3 - \frac{9}{4} - \frac{1}{2} (\ln 3)^2 \right] \right) \] \[ = \frac{1}{2} \left( \frac{25}{2} \ln 5 - \frac{25}{4} - \frac{1}{2} (\ln 5)^2 - \frac{9}{2} \ln 3 + \frac{9}{4} + \frac{1}{2} (\ln 3)^2 \right) \] \[ = \frac{1}{2} \left( \frac{25}{2}\ln 5 - \frac{9}{2}\ln 3 - \frac{25}{4} + \frac{9}{4} - \frac{1}{2} (\ln 5)^2 + \frac{1}{2} (\ln 3)^2 \right) \] \[ = \frac{1}{2} \left( \frac{25}{2}\ln 5 - \frac{9}{2}\ln 3 - 4 - \frac{1}{2} [(\ln 5)^2 - (\ln 3)^2] \right) \] \[ = \frac{25}{4}\ln 5 - \frac{9}{4}\ln 3 - 2 - \frac{1}{4} [(\ln 5)^2 - (\ln 3)^2] \] --- #### **Final answer for (b):** \[ \boxed{ \frac{25}{4}\ln 5 - \frac{9}{4}\ln 3 - 2 - \frac{1}{4} [(\ln 5)^2 - (\ln 3)^2] } \] --- ### (c) \(\int_0^1 \int_0^{x^3} xy \cos(3x) dy dx\) #### **Step 1: Integrate with respect to \(y\):** \[ \int_0^{x^3} xy \cos(3x) dy \] Note that \(x\) and \(\cos(3x)\) are constants with respect to \(y\): \[ = x \cos(3x) \int_0^{x^3} y dy = x \cos(3x) \left[ \frac{y^2}{2} \right]_0^{x^3} = x \cos(3x) \cdot \frac{(x^3)^2}{2} = x \cos(3x) \cdot \frac{x^6}{2} = \frac{x^7}{2} \cos(3x) \] --- #### **Step 2: Integrate with respect to \(x\):** \[ \int_0^1 \frac{x^7}{2} \cos(3x) dx = \frac{1}{2} \int_0^1 x^7 \cos(3x) dx \] This requires integration by parts: Let \(u = x^7\), \(dv = \cos(3x) dx\), \(du = 7x^6 dx\), \(v = \frac{\sin(3x)}{3}\): \[ \int x^7 \cos(3x) dx = x^7 \cdot \frac{\sin(3x)}{3} - \int \frac{\sin(3x)}{3} \cdot 7x^6 dx \] \[ = \frac{x^7 \sin(3x)}{3} - \frac{7}{3} \int x^6 \sin(3x) dx \] Now, \(\int x^6 \sin(3x) dx\) can be done by parts again, but this is a standard pattern, known as repeated integration by parts (tabular method or reduction formula). Let's denote: \[ I_n = \int x^n \cos(ax) dx \] The reduction formula is: \[ I_n = \frac{x^n \sin(ax)}{a} - \frac{n}{a} \int x^{n-1} \sin(ax) dx \] \[ J_n = \int x^n \sin(ax) dx = -\frac{x^n \cos(ax)}{a} + \frac{n}{a} \int x^{n-1} \cos(ax) dx \] So applying this recursively is quite time-consuming by hand, but for the purpose of this exercise, let's just write the final answer in terms of this reduction formula: So, \[ \int_0^1 x^7 \cos(3x) dx = \left[ \frac{x^7 \sin(3x)}{3} - \frac{7}{3} \int x^6 \sin(3x) dx \right]_0^1 \] Let’s compute \(\int x^6 \sin(3x) dx\) using the reduction formula above, but it's easier to use the general pattern or note that the final answer will be an expression involving sines and cosines evaluated at 0 and 1, with coefficients. --- #### **Final Answer for (c):** \[ \boxed{ \frac{1}{2} \int_0^1 x^7 \cos(3x) dx } \] Or, if you want the full reduction, it is: \[ \frac{1}{2} \left[ \frac{x^7 \sin(3x)}{3} + \frac{7x^6 \cos(3x)}{9} - \frac{42x^5 \sin(3x)}{27} - \frac{210x^4 \cos(3x)}{81} + \frac{840x^3 \sin(3x)}{243} + \frac{2520x^2 \cos(3x)}{729} - \frac{5040x \sin(3x)}{2187} - \frac{5040 \cos(3x)}{6561} \right]_0^1 \] Plug in \(x=1\) and \(x=0\): - At \(x=0\), all terms with \(x\) vanish, leaving only the last term: \(-\frac{5040}{6561} \cos(0) = -\frac{5040}{6561}(1)\) - At \(x=1\), substitute \(x=1\) into each term, with \(\sin(3)\) and \(\cos(3)\). So the explicit final answer is: \[ \frac{1}{2} \left\{ \left( \frac{\sin 3}{3} + \frac{7 \cos 3}{9} - \frac{42 \sin 3}{27} - \frac{210 \cos 3}{81} + \frac{840 \sin 3}{243} + \frac{2520 \cos 3}{729} - \frac{5040 \sin 3}{2187} - \frac{5040 \cos 3}{6561} \right) + \frac{5040}{6561} \right\} \] --- ## **Summary of Answers** **(

Let's carefully evaluate the corrected integrals step-by-step. --- ### **Part (b):** **Given integral:** \[ \int_{y=3}^{-4} \int_{x=1}^{5} \frac{x \ln y}{y} \, dx \, dy \] **Note**: The order of limits indicates \(y\) goes from 3 down to -4, which can be rewritten as: \[ \int_{y=-4}^{3} \int_{x=1}^{5} \frac{x \ln y}{y} \, dx \, dy \] **Step 1: Integrate with respect to \(x\):** \[ \int_{x=1}^{5} \frac{x \ln y}{y} dx \] Since \(\frac{\ln y}{y}\) does not depend on \(x\): \[ = \frac{\ln y}{y} \int_1^5 x \, dx = \frac{\ln y}{y} \left[ \frac{x^2}{2} \right]_1^5 = \frac{\ln y}{y} \left( \frac{25 - 1}{2} \right) = \frac{\ln y}{y} \cdot 12 \] So the inner integral simplifies to: \[ 12 \frac{\ln y}{y} \] --- ### **Step 2: Integrate with respect to \(y\):** \[ \int_{-4}^{3} 12 \frac{\ln y}{y} dy = 12 \int_{-4}^3 \frac{\ln y}{y} dy \] **Important note:** \(\ln y\) is only defined for \(y > 0\). The limits include negative \(y\), so the integral over negative \(y\) is invalid in the real domain. **Assumption:** The integral is meant over positive \(y\), i.e., from \(y=0\) to 3, or perhaps the original limits are different. But given the correction, it's from \(-4\) to 3, which crosses the domain of \(\ln y\). **Conclusion:** The integral over negative \(y\) is undefined for real numbers. To proceed meaningfully, we must restrict to \(y > 0\). Let's assume the integral is from \(0^+\) to 3: \[ \boxed{ \text{Integral} = 12 \int_0^3 \frac{\ln y}{y} dy } \] --- ### **Step 3: Evaluate \(\int_0^a \frac{\ln y}{y} dy\):** This is a standard integral: \[ \int_0^a \frac{\ln y}{y} dy = \frac{(\ln y)^2}{2} \bigg|_{y=0}^{a} \] At \(y=0^+\), \(\ln y \to -\infty\), so the integral diverges. Therefore, unless specified otherwise, the integral over \(0\) to \(a\) with the integrand \(\frac{\ln y}{y}\) diverges at 0. **Final note:** If the integral is from some positive lower limit \(a>0\), say from \(a=1\) to 3, then: \[ \int_1^{3} \frac{\ln y}{y} dy = \frac{(\ln y)^2}{2} \bigg|_1^{3} = \frac{(\ln 3)^2 - (\ln 1)^2}{2} = \frac{(\ln 3)^2 - 0}{2} = \frac{(\ln 3)^2}{2} \] and the total integral becomes: \[ 12 \times \frac{(\ln 3)^2}{2} = 6 (\ln 3)^2 \] --- ### **Final answer for Part (b):** \[ \boxed{ \textbf{If the limits are from } y=1 \text{ to } y=3: \quad 6 (\ln 3)^2 } \] --- ## **Part (c):** **Given integral:** \[ \int_{x=0}^{-} \int_{y=0}^{\pi/6} xy \cos(3x) \, dy \, dx \] **Interpreting limits:** - The outer integral is from \(x=0\) to some negative value (say, \(-a\)), or perhaps from \(-a\) to 0. Assuming the limits are from \(x=-a\) to 0, with \(a>0\), and the inner integral from \(y=0\) to \(\pi/6\). **Rewritten as:** \[ \int_{x=-a}^{0} \int_{y=0}^{\pi/6} xy \cos(3x) \, dy \, dx \] --- ### **Step 1: Integrate with respect to \(y\):** \[ \int_0^{\pi/6} xy \cos(3x) dy \] Since \(x\) and \(\cos(3x)\) are constants with respect to \(y\): \[ = x \cos(3x) \int_0^{\pi/6} y \, dy = x \cos(3x) \left[ \frac{y^2}{2} \right]_0^{\pi/6} = x \cos(3x) \cdot \frac{(\pi/6)^2}{2} \] \[ = x \cos(3x) \cdot \frac{\pi^2/36}{2} = x \cos(3x) \cdot \frac{\pi^2}{72} \] --- ### **Step 2: Integrate with respect to \(x\):** \[ \int_{x=-a}^{0} x \cos(3x) \, dx \] Note: The integral depends on \(a\). Let's assume \(a>0\). --- ### **Step 3: Evaluate \(\int x \cos(3x) dx\):** Use integration by parts: Let \(u = x\), \(dv = \cos(3x) dx\), then \(du = dx\), \(v = \frac{\sin(3x)}{3}\). \[ \int x \cos(3x) dx = x \cdot \frac{\sin(3x)}{3} - \int \frac{\sin(3x)}{3} dx \] \[ = \frac{x \sin(3x)}{3} + \frac{\cos(3x)}{9} + C \] --- ### **Step 4: Final evaluation:** \[ \int_{x=-a}^0 x \cos(3x) dx = \left[ \frac{x \sin(3x)}{3} + \frac{\cos(3x)}{9} \right]_{x=-a}^{x=0} \] At \(x=0\): \[ \frac{0 \cdot \sin 0}{3} + \frac{\cos 0}{9} = 0 + \frac{1}{9} = \frac{1}{9} \] At \(x=-a\): \[ \frac{-a \sin(-3a)}{3} + \frac{\cos(-3a)}{9} = - \frac{a \sin(-3a)}{3} + \frac{\cos(3a)}{9} \] But \(\sin(-\theta) = - \sin \theta\), so: \[ - \frac{a (-\sin 3a)}{3} + \frac{\cos 3a}{9} = \frac{a \sin 3a}{3} + \frac{\cos 3a}{9} \] Putting it all together: \[ \boxed{ \frac{1}{9} - \left( \frac{a \sin 3a}{3} + \frac{\cos 3a}{9} \right) = \frac{1}{9} - \frac{a \sin 3a}{3} - \frac{\cos 3a}{9} \] --- ### **Multiply by the factor from the y-integration:** Recall the inner integral result: \[ x \cos(3x) \cdot \frac{\pi^2}{72} \] So the full integral over \(x\): \[ \frac{\pi^2}{72} \left( \frac{1}{9} - \frac{a \sin 3a}{3} - \frac{\cos 3a}{9} \right) \] If \(a \to \infty\), the integral may diverge, but for finite \(a\), this is the explicit form. --- ## **Summary of final answers:** **(b):** \[ \boxed{ 6 (\ln 3)^2 \quad \text{(assuming limits from 1 to 3)} } \] **(c):** \[ \boxed{ \frac{\pi^2}{72} \left( \frac{1}{9} - \frac{a \sin 3a}{3} - \frac{\cos 3a}{9} \right) } \] --- **Note:** The precise numerical answer depends on the exact limits specified. If the limits are as assumed, these are the evaluated results.