Evaluate: \( \int e^{2z} \cos\left(\frac{1}{4}z\right) \, dz \)
Step 1: Let \( u = \cos\left(\frac{1}{4}z\right) \), then \( du = -\frac{1}{4} \sin\left(\frac{1}{4}z\right) dz \)
Let \( dv = e^{2z} dz \), then \( v = \frac{1}{2} e^{2z} \)
Step 2: Apply integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
\[ \int e^{2z} \cos\left(\frac{1}{4}z\right) \, dz = \frac{1}{2} e^{2z} \cos\left(\frac{1}{4}z\right) + \frac{1}{8} \int e^{2z} \sin\left(\frac{1}{4}z\right) \, dz \]
Step 3: Now integrate \( \int e^{2z} \sin\left(\frac{1}{4}z\right) \, dz \) using parts again.
Let \( u = \sin\left(\frac{1}{4}z\right) \), so \( du = \frac{1}{4} \cos\left(\frac{1}{4}z\right) dz \)
Let \( dv = e^{2z} dz \), so \( v = \frac{1}{2} e^{2z} \)
Step 4: Apply parts again:
\[ \int e^{2z} \sin\left(\frac{1}{4}z\right) \, dz = \frac{1}{2} e^{2z} \sin\left(\frac{1}{4}z\right) - \frac{1}{8} \int e^{2z} \cos\left(\frac{1}{4}z\right) \, dz \]
Step 5: Substitute back into original integral:
\[ I = \frac{1}{2} e^{2z} \cos\left(\frac{1}{4}z\right) + \frac{1}{8} \left( \frac{1}{2} e^{2z} \sin\left(\frac{1}{4}z\right) - \frac{1}{8} \int e^{2z} \cos\left(\frac{1}{4}z\right) \, dz \right) \]
Step 6: Simplify and solve for the integral:
\[ I = \frac{1}{2} e^{2z} \cos\left(\frac{1}{4}z\right) + \frac{1}{16} e^{2z} \sin\left(\frac{1}{4}z\right) - \frac{1}{64} I \]
Step 7: Solve for \( I \):
\[ I + \frac{1}{64} I = \frac{1}{2} e^{2z} \cos\left(\frac{1}{4}z\right) + \frac{1}{16} e^{2z} \sin\left(\frac{1}{4}z\right) \]
\[ \frac{65}{64} I = \frac{1}{2} e^{2z} \cos\left(\frac{1}{4}z\right) + \frac{1}{16} e^{2z} \sin\left(\frac{1}{4}z\right) \]
\[ I = \frac{64}{65} \left( \frac{1}{2} e^{2z} \cos\left(\frac{1}{4}z\right) + \frac{1}{16} e^{2z} \sin\left(\frac{1}{4}z\right) \right) + C \]
Evaluate: \( \int \frac{x^2 + 16}{x^4} \, dx \) using a trigonometric substitution.
Step 1: Recognize that the numerator \( x^2 + 16 \) resembles \( \tan^2\theta + 1 = \sec^2\theta \) if we set \( x = 4 \tan\theta \).
Step 2: Make the substitution:
Step 3: Substitute into the integral:
\[ \int \frac{x^2 + 16}{x^4} \, dx = \int \frac{(4 \tan\theta)^2 + 16}{(4 \tan\theta)^4} \cdot 4 \sec^2\theta \, d\theta \]
Simplify numerator and denominator:
\[ (4 \tan\theta)^2 + 16 = 16 \tan^2\theta + 16 = 16(\tan^2\theta + 1) = 16 \sec^2\theta \]
\[ (4 \tan\theta)^4 = 256 \tan^4\theta \]
So the integrand becomes:
\[ \int \frac{16 \sec^2\theta}{256 \tan^4\theta} \cdot 4 \sec^2\theta \, d\theta = \int \frac{64 \sec^4\theta}{256 \tan^4\theta} \, d\theta \]
Reduce constants:
\[ \frac{64}{256} = \frac{1}{4}, \quad \Rightarrow \int \frac{\sec^4\theta}{\tan^4\theta} \, d\theta \]
\[ \sec\theta = \frac{1}{\cos\theta}, \quad \tan\theta = \frac{\sin\theta}{\cos\theta} \]
\[ \Rightarrow \frac{\sec^4\theta}{\tan^4\theta} = \frac{1/\cos^4\theta}{\sin^4\theta/\cos^4\theta} = \frac{1}{\sin^4\theta} \]
So we now integrate:
\[ \frac{1}{4} \int \csc^4\theta \, d\theta \]
Use the standard reduction formula:
\[ \int \csc^4\theta \, d\theta = -\frac{1}{3} \csc^2\theta \cot\theta - \frac{2}{3} \ln|\csc\theta + \cot\theta| + C \]
So the result is:
\[ \frac{1}{4} \left( -\frac{1}{3} \csc^2\theta \cot\theta - \frac{2}{3} \ln|\csc\theta + \cot\theta| \right) + C \]
Recall: \( x = 4 \tan\theta \Rightarrow \tan\theta = \frac{x}{4} \)
From a right triangle:
Then:
Now plug back in:
\[ \int \frac{x^2 + 16}{x^4} \, dx = -\frac{1}{12} \cdot \frac{x^2 + 16}{x^2} \cdot \frac{4}{x} - \frac{1}{6} \ln\left|\frac{\sqrt{x^2 + 16} + 4}{x}\right| + C \]
Evaluate: \( \int \frac{3z^2 + 1}{(z+1)(z-5)^2} \, dz \)
Step 1: Set up the partial fractions decomposition:
\[ \frac{3z^2 + 1}{(z+1)(z-5)^2} = \frac{A}{z+1} + \frac{B}{z-5} + \frac{C}{(z-5)^2} \]
Step 2: Multiply both sides by \( (z+1)(z-5)^2 \) to eliminate denominators:
\[ 3z^2 + 1 = A(z-5)^2 + B(z+1)(z-5) + C(z+1) \]
Step 3: Expand the right-hand side:
\[ A(z^2 - 10z + 25) + B(z^2 - 4z - 5) + C(z + 1) \]
Combine like terms:
\[ (A + B)z^2 + (-10A - 4B + C)z + (25A - 5B + C) = 3z^2 + 0z + 1 \]
Step 4: Match coefficients:
Step 5: Solve the system of equations:
Now solve the system:
Step 6: Integrate each term:
\[ \int \left( \frac{1}{9(z+1)} + \frac{26}{9(z-5)} + \frac{38}{3(z-5)^2} \right) dz \]
Integrals:
Final Answer:
\[ \frac{1}{9} \ln|z+1| + \frac{26}{9} \ln|z-5| - \frac{38}{3(z-5)} + C \]
Evaluate: \( \lim_{z \to \infty} \frac{z^2 + e^{4z}}{2z - e^z} \)
Step 1: Check indeterminate form. As \( z \to \infty \), numerator and denominator → \( \infty \), so it's \( \frac{\infty}{\infty} \).
Step 2: Apply L'Hôpital's Rule (differentiate numerator and denominator):
\[ \frac{d}{dz}(z^2 + e^{4z}) = 2z + 4e^{4z}, \quad \frac{d}{dz}(2z - e^z) = 2 - e^z \]
So:
\[ \lim_{z \to \infty} \frac{2z + 4e^{4z}}{2 - e^z} \]
Step 3: As \( z \to \infty \), numerator → \( \infty \), denominator → \( -\infty \)
Answer: \( \lim = -\infty \)
Evaluate: \( \int_0^4 \frac{x}{x^2 - 9} \, dx \)
Step 1: Check for discontinuities. The denominator is 0 at \( x = 3 \in [0, 4] \). So the integral is improper.
Step 2: Split the integral at 3:
\[ \int_0^4 \frac{x}{x^2 - 9} \, dx = \int_0^3 \frac{x}{x^2 - 9} \, dx + \int_3^4 \frac{x}{x^2 - 9} \, dx \]
Step 3: Use substitution for both parts: let \( u = x^2 - 9 \Rightarrow du = 2x dx \Rightarrow \frac{du}{2} = x dx \)
So:
\[ \int \frac{x}{x^2 - 9} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|x^2 - 9| + C \]
Step 4: Apply limits:
\[ \lim_{a \to 3^-} \int_0^a \frac{x}{x^2 - 9} dx = \left[\frac{1}{2} \ln|x^2 - 9|\right]_0^a \]
As \( a \to 3 \), \( x^2 - 9 \to 0 \Rightarrow \ln|x^2 - 9| \to -\infty \)
Conclusion: The integral diverges.
Determine convergence or divergence of:
\[ \sum_{n=3}^{\infty} \frac{e^{-n}}{n^2 + 2n} \]
Step 1: Choose a comparison series. Since \( e^{-n} \) decays rapidly, and \( n^2 + 2n \sim n^2 \), compare with:
\[ b_n = \frac{e^{-n}}{n^2} \]
Step 2: Apply the Limit Comparison Test:
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{e^{-n}}{n^2 + 2n}}{\frac{e^{-n}}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n^2 + 2n} \]
Divide numerator and denominator by \( n^2 \):
\[ \lim_{n \to \infty} \frac{1}{1 + \frac{2}{n}} = 1 \]
Conclusion: Since the limit is a positive finite number and \( \sum \frac{e^{-n}}{n^2} \) converges (since \( e^{-n} \) decays faster than any polynomial), the given series converges by the Limit Comparison Test.
Test convergence of:
\[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{7 + 2n} \]
Step 1: Let \( a_n = \frac{1}{7 + 2n} \). This is positive and decreasing:
Conclusion: Alternating Series Test conditions are satisfied, so the series converges.
Absolute or Conditional?
Check \( \sum \left| \frac{(-1)^{n-1}}{7 + 2n} \right| = \sum \frac{1}{7 + 2n} \)
This behaves like \( \sum \frac{1}{2n} \sim \sum \frac{1}{n} \), which diverges.
Final Answer: The series converges conditionally, not absolutely.
Find the interval of convergence of:
\[ \sum_{n=0}^{\infty} \frac{4^{1 + 2n}}{5^n + 1} (x + 3)^n \]
Step 1: Use the Ratio Test:
\[ a_n = \frac{4^{1 + 2n}}{5^n + 1} (x + 3)^n \]
Compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \):
\[ = \lim_{n \to \infty} \left| \frac{4^{1 + 2(n+1)}}{5^{n+1} + 1} \cdot \frac{5^n + 1}{4^{1 + 2n}} \cdot (x + 3) \right| \]
Simplify exponential parts:
\[ \frac{4^{2n+3}}{4^{2n+1}} = 4^2 = 16 \]
So the limit becomes:
\[ 16 |x + 3| \cdot \lim_{n \to \infty} \frac{5^n + 1}{5^{n+1} + 1} = 16 |x + 3| \cdot \frac{1}{5} \]
\[ = \frac{16}{5} |x + 3| \]
Apply Ratio Test: the series converges when this limit is less than 1:
\[ \frac{16}{5} |x + 3| < 1 \Rightarrow |x + 3| < \frac{5}{16} \]
Step 2: Test endpoints \( x = -3 \pm \frac{5}{16} \)
Answer: The interval of convergence is \( \left( -3 - \frac{5}{16}, -3 + \frac{5}{16} \right) \) with possible endpoints pending further testing.
Find the power series for \( f(x) = \frac{7}{x^4} \) centered at \( c = -3 \)
Step 1: Rewrite function as: \( f(x) = 7(x - (-3))^{-4} = 7(x + 3)^{-4} \)
Step 2: Use generalized binomial series:
\[ (1 + u)^r = \sum_{n=0}^{\infty} \binom{r}{n} u^n, \quad \text{valid when } |u| < 1 \]
Let \( u = \frac{x + 3}{x + 3} - 1 = \left(1 - \frac{x + 3}{x + 3}\right) = 0 \), so instead differentiate known series:
Recall: \( \frac{1}{(1 - x)^k} = \sum_{n=0}^{\infty} \binom{n + k - 1}{n} x^n \)
So: \[ \frac{7}{(x + 3)^4} = 7 \sum_{n=0}^\infty \binom{n + 3}{3} (-1)^n (x + 3)^n \]
Answer: \[ \sum_{n=0}^{\infty} 7 \binom{n + 3}{3} (-1)^n (x + 3)^n \]
Given: \( x(t) = e^{-7t} + 2 \), \( y(t) = 6e^{2t} + e^{-3t} - 4t \)
Step 1: Compute \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)
\[ \frac{dy}{dx} = \frac{12e^{2t} - 3e^{-3t} - 4}{-7e^{-7t}} \]
Step 2: Compute \( \frac{d^2y}{dx^2} \)
\[ \frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{dy}{dx} \right) \div \frac{dx}{dt} \]
Use quotient rule and chain rule carefully; plug in \( t = -1 \) to compute slope and concavity numerically.
Given: \( r(\theta) = 2\sin(5\theta) \)
Find the area of one petal.
Step 1: Formula for polar area:
\[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2(\theta) \, d\theta \]
Each petal of \( \sin(n\theta) \) (for odd \( n \)) spans \( \theta = \frac{\pi}{n} \)
\[ A = \frac{1}{2} \int_0^{\frac{\pi}{5}} [2 \sin(5\theta)]^2 d\theta = 2 \int_0^{\frac{\pi}{5}} \sin^2(5\theta) d\theta \]
Use identity: \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \)
\[ A = 2 \int_0^{\frac{\pi}{5}} \frac{1 - \cos(10\theta)}{2} d\theta = \int_0^{\frac{\pi}{5}} (1 - \cos(10\theta)) d\theta \]
Integrate:
\[ \left[\theta - \frac{1}{10}\sin(10\theta)\right]_0^{\frac{\pi}{5}} = \frac{\pi}{5} \]
Final Answer: \( \frac{\pi}{5} \)
Given: \( r(\theta) = -4 \sin(\theta), \quad 0 \leq \theta \leq \pi \)
Step 1: Use arc length formula for polar curves:
\[ L = \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta \]
\[ r^2 = 16\sin^2\theta, \quad \left(\frac{dr}{d\theta}\right)^2 = 16\cos^2\theta \]
\[ L = \int_0^{\pi} \sqrt{16\sin^2\theta + 16\cos^2\theta} d\theta = \int_0^{\pi} \sqrt{16} \, d\theta = 4 \int_0^{\pi} d\theta = 4\pi \]
Final Answer: \( L = 4\pi \)