Let's analyze and compute each integral. --- ### **Integral A** \[ A = \int_0^1 x \sin^2(\pi x) dx \] Use the identity \(\sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta))\): \[ A = \int_0^1 x \cdot \frac{1}{2}(1 - \cos(2\pi x)) dx = \frac{1}{2} \int_0^1 x dx - \frac{1}{2} \int_0^1 x \cos(2\pi x) dx \] First integral: \[ \int_0^1 x dx = \frac{1}{2} \] Second integral (by parts, let \(u = x, dv = \cos(2\pi x) dx\)): - \(du = dx\) - \(v = \frac{1}{2\pi} \sin(2\pi x)\) \[ \int x \cos(2\pi x) dx = x \cdot \frac{1}{2\pi} \sin(2\pi x) - \int \frac{1}{2\pi} \sin(2\pi x) dx = \frac{x}{2\pi} \sin(2\pi x) + \frac{1}{(2\pi)^2} \cos(2\pi x) \] Evaluate from 0 to 1: - \(\sin(2\pi \cdot 1) = 0\), \(\sin(0) = 0\) - \(\cos(2\pi \cdot 1) = 1\), \(\cos(0) = 1\) So, \[ \left[ \frac{x}{2\pi} \sin(2\pi x) + \frac{1}{(2\pi)^2} \cos(2\pi x) \right]_0^1 = 0 + \frac{1}{4\pi^2}(1 - 1) = 0 \] So, \[ A = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} = 0.25 \] --- ### **Integral B** \[ B = \int_0^\pi \sin^3 x dx \] Use the reduction formula or rewrite using \(\sin^3 x = \frac{3\sin x - \sin(3x)}{4}\): \[ B = \int_0^\pi \frac{3\sin x - \sin(3x)}{4} dx = \frac{3}{4} \int_0^\pi \sin x dx - \frac{1}{4} \int_0^\pi \sin(3x) dx \] \[ \int_0^\pi \sin x dx = [-\cos x]_0^\pi = -\cos \pi + \cos 0 = 1 + 1 = 2 \] \[ \int_0^\pi \sin(3x) dx = [-\frac{1}{3}\cos(3x)]_0^\pi = -\frac{1}{3}[\cos(3\pi) - \cos(0)] = -\frac{1}{3}((-1) - 1) = -\frac{1}{3}(-2) = \frac{2}{3} \] So, \[ B = \frac{3}{4}(2) - \frac{1}{4}(\frac{2}{3}) = \frac{3}{2} - \frac{1}{6} = \frac{9}{6} - \frac{1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333 \] --- ### **Integral C** \[ C = \int_{-1}^1 (\sin(\pi x) + \cos(\pi x))^4 dx \] Let \(f(x) = (\sin(\pi x) + \cos(\pi x))^4\), note that it's an even function since \(\sin(\pi (-x)) = -\sin(\pi x)\) and \(\cos(\pi (-x)) = \cos(\pi x)\), so the sum is not odd or even, but let's expand: Given: \((a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\) So, \[ (\sin(\pi x) + \cos(\pi x))^4 = \sin^4(\pi x) + 4 \sin^3(\pi x) \cos(\pi x) + 6 \sin^2(\pi x) \cos^2(\pi x) + 4 \sin(\pi x) \cos^3(\pi x) + \cos^4(\pi x) \] Now integrate from \(-1\) to \(1\). But notice that the integrand is symmetric around \(x=0\), and over one period (\(-1\) to \(1\)), the average values of sine and cosine powers are the same. This integral will be positive. But let's quickly estimate it: - The terms with odd powers of sine/cosine will integrate to 0 because they're odd functions over \([-1,1]\). - Only the even powers remain. So, \[ C = \int_{-1}^1 [\sin^4(\pi x) + 6\sin^2(\pi x)\cos^2(\pi x) + \cos^4(\pi x)] dx \] But \(\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta\) and \(\sin^2\theta \cos^2\theta = \frac{1}{4}\sin^2(2\theta)\), and the average value over a period is \(1/2\) for \(\sin^2\) and \(\cos^2\). If you work out the details, this integral evaluates to 3 (since the period is 2 and the mean value of the function over the period is 1.5). --- ### **Integral D** \[ D = \int_{-\pi}^{\pi} x \cos^3 x dx \] \(x\) is an odd function, \(\cos^3 x\) is an odd function (since cosine is even and the cube of an even function is even), so their product is odd. The integral of an odd function over a symmetric interval about zero is **zero**. **So \(D = 0\).** --- ### **Integral E** \[ E = \int_{-\pi}^{\pi} x \sin x \cos^3 x dx \] \(x\) is odd, \(\sin x\) is odd, \(\cos^3 x\) is even, so their product is odd. The integral of an odd function over a symmetric interval about zero is **zero**. **So \(E = 0\).** --- ## **Arranging in Increasing Order** - \(D = 0\) - \(E = 0\) - \(A = 0.25\) - \(B = 1.333\) - \(C = 3\) ### **Final Order:** 1. \(D\) 2. \(E\) 3. \(A\) 4. \(B\) 5. \(C\) --- ### **Fill in the blanks:** 1. \(D\) 2. \(E\) 3. \(A\) 4. \(B\) 5. \(C\) --- **If you want, you can swap D and E in positions 1 and 2, since both are 0.**

Let's carefully analyze and compute each integral step-by-step, providing detailed explanations. --- ### **Integral A** \[ A = \int_0^1 x \sin^2(\pi x) \, dx \] **Step 1:** Use the identity for \(\sin^2 \theta\): \[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \] So, \[ A = \int_0^1 x \cdot \frac{1 - \cos 2\pi x}{2} dx = \frac{1}{2} \int_0^1 x dx - \frac{1}{2} \int_0^1 x \cos 2\pi x dx \] **Step 2:** Compute \(\int_0^1 x dx\): \[ \int_0^1 x dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2} \] **Step 3:** Compute \(\int_0^1 x \cos 2\pi x dx\): Use **integration by parts**: - Let \(u = x \Rightarrow du = dx\) - Let \(dv = \cos 2\pi x dx \Rightarrow v = \frac{1}{2\pi} \sin 2\pi x\) Applying integration by parts: \[ \int x \cos 2\pi x dx = uv - \int v du = x \cdot \frac{1}{2\pi} \sin 2\pi x - \int \frac{1}{2\pi} \sin 2\pi x dx \] Compute the second integral: \[ \int \sin 2\pi x dx = -\frac{1}{2\pi} \cos 2\pi x \] Putting it all together: \[ \int x \cos 2\pi x dx = \frac{x}{2\pi} \sin 2\pi x + \frac{1}{(2\pi)^2} \cos 2\pi x + C \] Evaluate from 0 to 1: At \(x=1\): - \(\sin 2\pi \cdot 1 = 0\) - \(\cos 2\pi \cdot 1 = 1\) At \(x=0\): - \(\sin 0 = 0\) - \(\cos 0 = 1\) Thus: \[ \left[\frac{x}{2\pi} \sin 2\pi x + \frac{1}{(2\pi)^2} \cos 2\pi x \right]_0^1 = \left( 0 + \frac{1}{4\pi^2} \cdot 1 \right) - \left( 0 + \frac{1}{4\pi^2} \cdot 1 \right) = 0 \] **Final value:** \[ \int_0^1 x \cos 2\pi x dx = 0 \] **Step 4:** Assemble the parts: \[ A = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot 0 = \frac{1}{4} \] **Answer for A:** \[ \boxed{A = 0.25} \] --- ### **Integral B** \[ B = \int_0^\pi \sin^3 x \, dx \] **Step 1:** Use the reduction formula: \[ \sin^3 x = \sin x \cdot \sin^2 x = \sin x \left( \frac{1 - \cos 2x}{2} \right) \] So, \[ B = \int_0^\pi \sin x \cdot \frac{1 - \cos 2x}{2} dx = \frac{1}{2} \int_0^\pi \sin x dx - \frac{1}{2} \int_0^\pi \sin x \cos 2x dx \] **Step 2:** Compute \(\int_0^\pi \sin x dx\): \[ \left[ -\cos x \right]_0^\pi = -\cos \pi + \cos 0 = -(-1) + 1 = 2 \] **Step 3:** Compute \(\int_0^\pi \sin x \cos 2x dx\): Use the product-to-sum formula: \[ \sin x \cos 2x = \frac{1}{2} [ \sin (x + 2x) + \sin (x - 2x)] = \frac{1}{2} [ \sin 3x + \sin (-x)] = \frac{1}{2} [ \sin 3x - \sin x] \] Thus, \[ \int_0^\pi \sin x \cos 2x dx = \frac{1}{2} \int_0^\pi \sin 3x dx - \frac{1}{2} \int_0^\pi \sin x dx \] Calculate each: - \(\int_0^\pi \sin 3x dx = \left[ -\frac{1}{3} \cos 3x \right]_0^\pi = -\frac{1}{3} (\cos 3\pi - \cos 0) = -\frac{1}{3}((-1) - 1) = -\frac{1}{3}(-2) = \frac{2}{3}\) - \(\int_0^\pi \sin x dx = 2\) (from earlier) Putting together: \[ \int_0^\pi \sin x \cos 2x dx = \frac{1}{2} \cdot \frac{2}{3} - \frac{1}{2} \cdot 2 = \frac{1}{3} - 1 = -\frac{2}{3} \] **Step 4:** Final calculation: \[ B = \frac{1}{2} \times 2 - \frac{1}{2} \times \left(-\frac{2}{3}\right) = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.333 \] **Answer for B:** \[ \boxed{B = \frac{4}{3} \approx 1.333} \] --- ### **Integral C** \[ C = \int_{-1}^1 (\sin \pi x + \cos \pi x)^4 dx \] **Step 1:** Expand using binomial theorem: \[ (\sin \pi x + \cos \pi x)^4 = \sin^4 \pi x + 4 \sin^3 \pi x \cos \pi x + 6 \sin^2 \pi x \cos^2 \pi x + 4 \sin \pi x \cos^3 \pi x + \cos^4 \pi x \] **Step 2:** Identify symmetry: - \(\sin \pi x\): odd function - \(\cos \pi x\): even function Terms: - \(\sin^4 \pi x\): even - \(4 \sin^3 \pi x \cos \pi x\): odd (since \(\sin^3\) is odd, multiplied by even \(\cos\)) - \(6 \sin^2 \pi x \cos^2 \pi x\): even - \(4 \sin \pi x \cos^3 \pi x\): odd - \(\cos^4 \pi x\): even **Step 3:** Integrate only the even functions: \[ C = \int_{-1}^1 \left[\sin^4 \pi x + 6 \sin^2 \pi x \cos^2 \pi x + \cos^4 \pi x \right] dx \] Because the odd terms integrate to zero over symmetric bounds. --- ### **Step 4:** Simplify the sum of even terms: Recall identities: \[ \sin^4 \theta = \frac{3 - 4 \cos 2\theta + \cos 4\theta}{8} \] \[ \cos^4 \theta = \frac{3 + 4 \cos 2\theta + \cos 4\theta}{8} \] \[ \sin^2 \theta \cos^2 \theta = \frac{\sin^2 2\theta}{4} = \frac{1 - \cos 4\theta}{8} \] Substitute: \[ C = \int_{-1}^1 \left[ \frac{3 - 4 \cos 2\pi x + \cos 4\pi x}{8} + 6 \times \frac{1 - \cos 4\pi x}{8} + \frac{3 + 4 \cos 2\pi x + \cos 4\pi x}{8} \right] dx \] Simplify numerator: \[ \left[ \frac{3 - 4 \cos 2\pi x + \cos 4\pi x + 6(1 - \cos 4\pi x) + 3 + 4 \cos 2\pi x + \cos 4\pi x }{8} \right] \] Combine like terms: - Constants: \(3 + 6 + 3 = 12\) - \(\cos 2\pi x\): \(-4 + 4 = 0\) - \(\cos 4\pi x\): \(1 - 6 + 1 = -4\) Total numerator: \[ 12 + 0 \times \cos 2\pi x - 4 \cos 4\pi x \] Thus, \[ C = \int_{-1}^1 \frac{12 - 4 \cos 4\pi x}{8} dx = \frac{1}{8} \int_{-1}^1 12 dx - \frac{1}{8} \int_{-1}^1 4 \cos 4\pi x dx \] Compute: \[ \int_{-1}^1 dx = 2 \] \[ \int_{-1}^1 \cos 4\pi x dx = \left[ \frac{\sin 4\pi x}{4\pi} \right]_{-1}^1 = \frac{\sin 4\pi \cdot 1 - \sin (-4\pi)}{4\pi} \] But \(\sin 4\pi = 0\), so numerator = 0. Therefore, \[ C = \frac{1}{8} \times 12 \times 2 - 0 = \frac{12}{8} \times 2 = \frac{12}{8} \times 2 = \frac{12 \times 2}{8} = \frac{24}{8} = 3 \] **Answer for C:** \[ \boxed{C = 3} \] --- ### **Integral D** \[ D = \int_{-\pi}^\pi x \cos^3 x dx \] **Step 1:** Note the parity: - \(x\) is odd - \(\cos^3 x\): even (since \(\cos x\) is even, cube preserves evenness) Product: \[ x \times \cos^3 x = \text{odd} \times \text{even} = \text{odd} \] **Step 2:** Integrate over symmetric bounds: \[ \int_{-a}^a \text{(odd function)} dx = 0 \] Thus, \[ D = 0 \] --- ### **Integral E** \[ E = \int_{-\pi}^\pi x \sin x \cos^3 x dx \] **Step 1:** Parity considerations: - \(x\): odd - \(\sin x\): odd - \(\cos^3 x\): even Product: \[ x \times \sin x \times \cos^3 x = (\text{odd}) \times (\text{odd}) \times (\text{even}) = (\text{even}) \times (\text{even}) = \text{even} \] - because odd \(\times\) odd = even, and even \(\times\) even = even. **Step 2:** The integral over symmetric bounds: \[ \int_{-a}^a \text{(even)} dx = 2 \times \int_0^a \text{(even)} dx \] Thus, the integral is **not zero**; we need explicit calculation. --- ### **Step 3:** Simplify the integrand: Use the identity: \[ \sin x \cos^3 x = \sin x (\cos^2 x)^1 \times \cos x \] Alternatively, express: \[ \sin x \cos^3 x = \frac{1}{4} \sin x (3 + \cos 2x) \times \cos x \] But easier is to write \(\sin x \cos^3 x\) as: \[ \sin x \cos^3 x = \frac{1}{4} \sin x (3 + \cos 2x) \quad \text{(using power reduction formulas)} \] Alternatively, use the identity directly: \[ \sin x \cos^3 x = \frac{1}{4} \left(3 \sin x + \sin

Apologies for the interruption. Continuing with the detailed calculation for integral **E**: --- ### **Integral E** \[ E = \int_{-\pi}^\pi x \sin x \cos^3 x \, dx \] **Step 1:** Parity analysis: - \(x\): odd - \(\sin x\): odd - \(\cos^3 x\): even Since odd \(\times\) odd = even, and even \(\times\) even = even, the overall integrand: \[ x \times \sin x \times \cos^3 x \] is **even**. Therefore, the integral over \([- \pi, \pi]\): \[ E = 2 \int_0^\pi x \sin x \cos^3 x \, dx \] **Step 2:** Simplify the integrand: Express \(\cos^3 x\) using power reduction: \[ \cos^3 x = \cos x \times \cos^2 x \] And recall: \[ \cos^2 x = \frac{1 + \cos 2x}{2} \] So, \[ \cos^3 x = \cos x \times \frac{1 + \cos 2x}{2} = \frac{1}{2} \cos x + \frac{1}{2} \cos x \cos 2x \] Next, express \(\cos x \cos 2x\): \[ \cos x \cos 2x = \frac{1}{2} [ \cos (x - 2x) + \cos (x + 2x) ] = \frac{1}{2} [ \cos (-x) + \cos 3x ] = \frac{1}{2} [ \cos x + \cos 3x ] \] Therefore, \[ \cos^3 x = \frac{1}{2} \cos x + \frac{1}{2} \times \frac{1}{2} [ \cos x + \cos 3x ] = \frac{1}{2} \cos x + \frac{1}{4} \cos x + \frac{1}{4} \cos 3x \] Combine like terms: \[ \cos^3 x = \left( \frac{1}{2} + \frac{1}{4} \right) \cos x + \frac{1}{4} \cos 3x = \frac{3}{4} \cos x + \frac{1}{4} \cos 3x \] **Step 3:** Substitute back into the integrand: \[ x \sin x \cos^3 x = x \sin x \left( \frac{3}{4} \cos x + \frac{1}{4} \cos 3x \right) \] \[ = \frac{3}{4} x \sin x \cos x + \frac{1}{4} x \sin x \cos 3x \] **Step 4:** Use product-to-sum identities: - \(\sin x \cos x = \frac{1}{2} \sin 2x\) - \(\sin x \cos 3x = \frac{1}{2} [ \sin (x + 3x) + \sin (x - 3x) ] = \frac{1}{2} [ \sin 4x + \sin (-2x) ] = \frac{1}{2} (\sin 4x - \sin 2x)\) So, \[ x \sin x \cos^3 x = \frac{3}{4} x \times \frac{1}{2} \sin 2x + \frac{1}{4} x \times \frac{1}{2} (\sin 4x - \sin 2x) \] \[ = \frac{3}{8} x \sin 2x + \frac{1}{8} x \sin 4x - \frac{1}{8} x \sin 2x \] Combine the \(\sin 2x\) terms: \[ \left( \frac{3}{8} - \frac{1}{8} \right) x \sin 2x = \frac{2}{8} x \sin 2x = \frac{1}{4} x \sin 2x \] Thus, \[ x \sin x \cos^3 x = \frac{1}{4} x \sin 2x + \frac{1}{8} x \sin 4x \] **Step 5:** Final integral: \[ E = 2 \times \int_0^\pi \left( \frac{1}{4} x \sin 2x + \frac{1}{8} x \sin 4x \right) dx = \frac{1}{2} \int_0^\pi x \sin 2x dx + \frac{1}{4} \int_0^\pi x \sin 4x dx \] --- ### **Step 6:** Compute \(\int_0^\pi x \sin kx dx\): Use **integration by parts**: Let \(u = x \Rightarrow du = dx\) \(dv = \sin kx dx \Rightarrow v = -\frac{1}{k} \cos kx\) Therefore, \[ \int x \sin kx dx = -\frac{x}{k} \cos kx + \frac{1}{k} \int \cos kx dx = - \frac{x}{k} \cos kx + \frac{1}{k^2} \sin kx + C \] Evaluate from 0 to \(\pi\): \[ \int_0^\pi x \sin kx dx = \left[ - \frac{x}{k} \cos kx + \frac{1}{k^2} \sin kx \right]_0^\pi \] At \(x=\pi\): - \(\cos k\pi = (-1)^k\) - \(\sin k \pi = 0\) At \(x=0\): - \(\cos 0 = 1\) - \(\sin 0 = 0\) Thus, \[ \int_0^\pi x \sin kx dx = - \frac{\pi}{k} (-1)^k + 0 - (0 + 0) = - \frac{\pi}{k} (-1)^k \] --- ### **Step 7:** Compute the two integrals: \[ \int_0^\pi x \sin 2x dx = - \frac{\pi}{2} (-1)^2 = - \frac{\pi}{2} \times 1 = - \frac{\pi}{2} \] \[ \int_0^\pi x \sin 4x dx = - \frac{\pi}{4} (-1)^4 = - \frac{\pi}{4} \times 1 = - \frac{\pi}{4} \] --- ### **Step 8:** Final calculation: \[ E = \frac{1}{2} \times \left( - \frac{\pi}{2} \right) + \frac{1}{4} \times \left( - \frac{\pi}{4} \right) = - \frac{\pi}{4} - \frac{\pi}{16} = - \left( \frac{4\pi}{16} + \frac{\pi}{16} \right) = - \frac{5\pi}{16} \] --- ### **Final answer for E:** \[ \boxed{E = - \frac{5\pi}{16} \approx -0.9817} \] --- ### **Summary of all integrals:** | Integral | Value | Approximate (if needed) | |------------|--------|-------------------------| | **A** | 0.25 | 0.25 | | **B** | 4/3 | 1.333 | | **C** | 3 | 3.00 | | **D** | 0 | 0 | | **E** | \(-\frac{5\pi}{16}\) | \(-0.9817\) | --- ### **Arranged in increasing order:** 1. \(D = 0\) 2. \(E \approx -0.9817\) 3. \(A = 0.25\) 4. \(B \approx 1.333\) 5. \(C = 3\) --- **Final answer:** \[ \boxed{ \text{Order: } D, E, A, B, C } \] --- If you need any further clarification, feel free to ask!