Let's solve **C** step-by-step: ### C. \(\int_0^1 x^4 (1-x)^3 dx\) This is a Beta function integral of the form: \[ \int_0^1 x^{m-1}(1-x)^{n-1} dx = B(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \] Here, \(x^4 = x^{5-1}\) and \((1-x)^3 = (1-x)^{4-1}\), so \(m = 5\), \(n = 4\). So, \[ \int_0^1 x^4 (1-x)^3 dx = B(5,4) = \frac{\Gamma(5)\Gamma(4)}{\Gamma(5+4)} \] Now evaluate the Gamma functions: - \(\Gamma(5) = 4! = 24\) - \(\Gamma(4) = 3! = 6\) - \(\Gamma(9) = 8! = 40320\) So, \[ B(5,4) = \frac{24 \times 6}{40320} = \frac{144}{40320} = \frac{1}{280} \] But the provided answer is \(\frac{1}{5}\). Let's check the calculation: Wait, let's re-express the integral: \[ \int_0^1 x^4 (1-x)^3 dx = \frac{\Gamma(5)\Gamma(4)}{\Gamma(9)} \] From the calculation above, \(\frac{144}{40320} = \frac{1}{280}\). But the image says the answer is \(\frac{1}{5}\), which seems off based on the Beta function. Let me check using the binomial expansion: Expand \((1-x)^3 = 1 - 3x + 3x^2 - x^3\): So, \[ \int_0^1 x^4 (1-x)^3 dx = \int_0^1 x^4 (1 - 3x + 3x^2 - x^3) dx \] \[ = \int_0^1 x^4 dx - 3\int_0^1 x^5 dx + 3\int_0^1 x^6 dx - \int_0^1 x^7 dx \] \[ = \left[\frac{x^5}{5}\right]_0^1 - 3\left[\frac{x^6}{6}\right]_0^1 + 3\left[\frac{x^7}{7}\right]_0^1 - \left[\frac{x^8}{8}\right]_0^1 \] \[ = \frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8} \] \[ = \frac{1}{5} - \frac{1}{2} + \frac{3}{7} - \frac{1}{8} \] Now, find common denominator (280): \[ = \frac{56}{280} - \frac{140}{280} + \frac{120}{280} - \frac{35}{280} \] \[ = (56 - 140 + 120 - 35)/280 = (56 + 120) - (140 + 35) = 176 - 175 = 1 \] \[ \frac{1}{280} \] So the correct answer is \(\boxed{\frac{1}{280}}\), not \(\frac{1}{5}\) as shown in the image. The image appears to have a typo. --- ### **Final Answer for C:** \[ \boxed{\frac{1}{280}} \]

Certainly! Let's carefully evaluate each part step-by-step with detailed calculations and explanations. --- ### **Part C: \(\int_0^1 x^4 (1 - x)^3 dx\)** This is a Beta integral, which has the form: \[ \int_0^1 x^{m-1} (1 - x)^{n-1} dx = B(m, n) = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m + n)} \] **Identify parameters:** - \(x^4 = x^{5-1} \Rightarrow m = 5\) - \((1 - x)^3 = (1 - x)^{4-1} \Rightarrow n = 4\) **Apply Beta function:** \[ \int_0^1 x^4 (1 - x)^3 dx = B(5, 4) = \frac{\Gamma(5) \Gamma(4)}{\Gamma(9)} \] **Calculate Gamma functions:** - \(\Gamma(5) = 4! = 24\) - \(\Gamma(4) = 3! = 6\) - \(\Gamma(9) = 8! = 40320\) **Compute the integral:** \[ \frac{24 \times 6}{40320} = \frac{144}{40320} \] Simplify numerator and denominator: \[ \frac{144}{40320} \] Divide numerator and denominator by 144: \[ \frac{1}{280} \] **Thus, the value of the integral is:** \[ \boxed{\frac{1}{280}} \] --- ### **Summary for Part C:** \[ \boxed{ \int_0^1 x^4 (1 - x)^3 dx = \frac{1}{280} } \] --- ### **Note:** The image's provided answer appears to be \(\frac{1}{5}\), but based on the correct Beta function calculation and expansion, the accurate result is \(\frac{1}{280}\). --- Let me know if you'd like me to analyze the other parts of the question!

Certainly! Let's evaluate each part step-by-step with detailed explanations. --- ## Part A: **Expression:** \[ \int_{SE} x \, \text{Tex} \, dx - \int_{Jo} x e^{mx} dx \quad \text{(as per the image, but notation is unclear)} \] **Note:** The image has some unclear notation, but based on typical integral problems, I will interpret and evaluate the integrals that are likely intended. --- ## Part B: **Expression:** \[ \int_{ps} x^2 e^{-mx} dx + \int_{1} x e dx \] --- ## Part C: **Expression:** \[ \int_0^1 x^4 (1 - x)^3 dx \] **Evaluation (detailed below):** --- ## Part D: **Expression:** \[ \int \sin^3 \theta \cos^6 \theta \, d\theta \] --- ## Part E: **Show that:** \[ \int \sin^n \theta \cos^m \theta \, d\theta \] --- Now, let's evaluate each part carefully. --- # Part C (Detailed Calculation): \[ \boxed{ \int_0^1 x^4 (1 - x)^3 dx } \] ### Step 1: Recognize the integral as a Beta function \[ \int_0^1 x^{m-1} (1 - x)^{n-1} dx = B(m, n) = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m + n)} \] ### Step 2: Identify \(m\) and \(n\): \[ x^4 = x^{5-1} \Rightarrow m = 5 \] \[ (1 - x)^3 = (1 - x)^{4-1} \Rightarrow n = 4 \] ### Step 3: Write the integral as Beta function: \[ \int_0^1 x^4 (1 - x)^3 dx = B(5, 4) = \frac{\Gamma(5) \Gamma(4)}{\Gamma(9)} \] ### Step 4: Compute Gamma functions: \[ \Gamma(5) = 4! = 24 \] \[ \Gamma(4) = 3! = 6 \] \[ \Gamma(9) = 8! = 40320 \] ### Step 5: Calculate the value: \[ \frac{24 \times 6}{40320} = \frac{144}{40320} \] Divide numerator and denominator by 144: \[ \frac{1}{280} \] ### **Final answer for Part C:** \[ \boxed{ \int_0^1 x^4 (1 - x)^3 dx = \frac{1}{280} } \] --- # Part D: \[ I = \int \sin^3 \theta \cos^6 \theta \, d\theta \] ### Step 1: Use identities and substitution Express \(\sin^3 \theta\) as: \[ \sin^3 \theta = \sin \theta \cdot \sin^2 \theta \] Recall: \[ \sin^2 \theta = 1 - \cos^2 \theta \] Thus, \[ I = \int \sin \theta (1 - \cos^2 \theta) \cos^6 \theta \, d\theta \] ### Step 2: Substitution Let: \[ u = \cos \theta \Rightarrow du = -\sin \theta d\theta \] Rearranged: \[ \sin \theta d\theta = -du \] Express the integral in \(u\): \[ I = - \int (1 - u^2) u^6 du \] ### Step 3: Expand the integrand \[ I = - \int (u^6 - u^8) du \] ### Step 4: Integrate term by term \[ I = - \left( \frac{u^7}{7} - \frac{u^9}{9} \right) + C \] ### Step 5: Back-substitute \(u = \cos \theta\) \[ I = - \left( \frac{\cos^7 \theta}{7} - \frac{\cos^9 \theta}{9} \right) + C \] --- ## **Final expression:** \[ \boxed{ \int \sin^3 \theta \cos^6 \theta \, d\theta = - \left( \frac{\cos^7 \theta}{7} - \frac{\cos^9 \theta}{9} \right) + C } \] --- # Part E: **Show that:** \[ \int \sin^n \theta \cos^m \theta \, d\theta \] **can be expressed in terms of powers of \(\sin \theta\) or \(\cos \theta\).** --- ## General approach: - For **odd powers**, express one factor using the identity: \[ \sin^n \theta = \sin^{n-1} \theta \cdot \sin \theta \] then use substitution \(u = \cos \theta\) or \(u= \sin \theta\). - For **even powers**, use power-reduction formulas: \[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \] \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \] --- ## **Example:** \[ \int \sin^n \theta \cos^m \theta d\theta \] - If \(n\) is odd, write: \[ \sin^n \theta = \sin^{n-1} \theta \cdot \sin \theta \] - Use substitution: \[ u = \cos \theta, \quad du = - \sin \theta d\theta \] - Rewrite the integral in terms of \(u\): \[ \int \sin^{n-1} \theta \cdot \sin \theta \cos^m \theta d\theta = - \int (1 - u^2)^{\frac{n-1}{2}} u^m du \] which can then be evaluated. --- ## **Summary:** - Use identities and substitution based on the parity of powers. - For even powers, use power reduction formulas. - For odd powers, factor out \(\sin \theta\) or \(\cos \theta\) and substitute. --- # **Final notes:** - The key to integrating powers of sine and cosine is to reduce the powers using identities and substitution. - Beta and Gamma functions are useful for integrals over \([0,1]\) with polynomial factors. --- Let me know if you'd like me to clarify any specific part further!