Step-by-Step Calculus Exam Guidance: SF1625 Envariabelanalys

Q1a. Beräkna integralen $\int_0^1 \frac{x-7}{x^2-2x-3} \, dx$

Background

Topic: Integration Techniques (Rational Functions)

This question tests your ability to integrate a rational function, which often involves partial fraction decomposition.

Key Terms and Formulas:

• Partial fraction decomposition: Expressing a rational function as a sum of simpler fractions.

• Definite integral: $\int_a^b f(x) \, dx$ calculates the area under $f(x)$ from $x=a$ to $x=b$.

Step-by-Step Guidance

1. Factor the denominator: $x^2 - 2x - 3 = (x - 3)(x + 1)$.

2. Rewrite the integrand using partial fractions: $\frac{x-7}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1}$, and solve for $A$ and $B$.

3. Set up the integral as the sum of two simpler integrals: $\int_0^1 \left( \frac{A}{x-3} + \frac{B}{x+1} \right) dx$.

4. Recall the integral formula: $\int \frac{1}{x-a} dx = \ln|x-a| + C$.

Try solving on your own before revealing the answer!

Final Answer:

After partial fraction decomposition, $A = \frac{1}{2}$ and $B = -\frac{3}{2}$.

The integral becomes $\int_0^1 \left( \frac{1}{2(x-3)} - \frac{3}{2(x+1)} \right) dx$.

Evaluating, the final answer is $\frac{1}{2} \ln|x-3| - \frac{3}{2} \ln|x+1|$ from $x=0$ to $x=1$, which gives:

$\left[ \frac{1}{2} \ln|x-3| - \frac{3}{2} \ln|x+1| \right]_0^1 = \frac{1}{2} \ln 2 - \frac{3}{2} \ln 2 - \frac{1}{2} \ln 3 + \frac{3}{2} \ln 1 = -\ln 2 - \frac{1}{2} \ln 3$

So, the final answer is $-\ln 2 - \frac{1}{2} \ln 3$.

Q1b. Beräkna integralen $\int_0^2 \frac{x^2}{4 + x^2} \, dx$

Background

Topic: Integration Techniques (Substitution)

This question tests your ability to integrate a rational function, possibly using substitution.

Key Terms and Formulas:

• Substitution: Letting $u = x^2$ or $u = 4 + x^2$ to simplify the integral.

• Integral formula: $\int \frac{x^2}{a^2 + x^2} dx$ can be solved by splitting or substitution.

Step-by-Step Guidance

1. Rewrite the integrand: $\frac{x^2}{4 + x^2} = 1 - \frac{4}{4 + x^2}$.

2. Split the integral: $\int_0^2 \left(1 - \frac{4}{4 + x^2}\right) dx = \int_0^2 1 dx - 4 \int_0^2 \frac{1}{4 + x^2} dx$.

3. Recall the formula: $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$.

4. Set up the definite integrals for each part, but do not evaluate yet.

Try solving on your own before revealing the answer!

Final Answer:

The integral evaluates to $2 - 4 \cdot \frac{1}{2} \left[ \arctan\left(\frac{x}{2}\right) \right]_0^2$.

Plugging in the limits, $\arctan(1) = \frac{\pi}{4}$ and $\arctan(0) = 0$.

So, the final answer is $2 - 2 \cdot \frac{\pi}{4} = 2 - \frac{\pi}{2}$.

Q2a. Låt $f(x) = (x^2 - 4x + 4) e^x$. Bestäm gränsvärdena $\lim_{x \to -\infty} f(x)$ och $\lim_{x \to \infty} f(x)$.

Background

Topic: Limits and Asymptotic Behavior

This question tests your understanding of how functions behave as $x$ approaches infinity or negative infinity.

Key Terms and Formulas:

• Limit: $\lim_{x \to a} f(x)$ describes the value $f(x)$ approaches as $x$ gets close to $a$.

• Exponential growth/decay: $e^x$ grows rapidly as $x \to \infty$, decays as $x \to -\infty$.

Step-by-Step Guidance

1. Rewrite $f(x)$: $f(x) = (x^2 - 4x + 4) e^x$.

2. Analyze the behavior as $x \to \infty$: The polynomial part grows, but $e^x$ grows much faster.

3. Analyze as $x \to -\infty$: $e^x$ approaches zero, while the polynomial part grows large (positive or negative).

4. Set up the limits for each case, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

$\lim_{x \to -\infty} f(x) = 0$ (since $e^x \to 0$ faster than the polynomial grows).

$\lim_{x \to \infty} f(x) = \infty$ (since both the polynomial and $e^x$ grow, and $e^x$ dominates).

Q2b. Avgör på vilka intervall $f$ är växande respektive avtagande.

Background

Topic: Increasing/Decreasing Functions (First Derivative Test)

This question tests your ability to use derivatives to determine where a function is increasing or decreasing.

Key Terms and Formulas:

• First derivative: $f'(x)$ tells you where $f$ is increasing ($f'(x) > 0$) or decreasing ($f'(x) < 0$).

• Product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.

Step-by-Step Guidance

1. Compute $f'(x)$ using the product rule: $f'(x) = \frac{d}{dx}[(x^2 - 4x + 4) e^x]$.

2. Set $f'(x) = 0$ to find critical points.

3. Analyze the sign of $f'(x)$ in the intervals determined by the critical points.

4. Determine which intervals correspond to increasing or decreasing behavior, but do not list them yet.

Try solving on your own before revealing the answer!

Final Answer:

$f'(x) = (x^2 - 4x + 4) e^x + (2x - 4) e^x = (x^2 - 2x) e^x$.

Set $x^2 - 2x = 0 \implies x = 0$ or $x = 2$.

$f$ is increasing for $x > 2$ and $x < 0$, decreasing for $0 < x < 2$.

Q2c. Bestäm alla lokala extremvärden till $f$ och skissa grafen till $f$.

Background

Topic: Local Extrema (Critical Points)

This question tests your ability to find local maxima and minima using derivatives.

Key Terms and Formulas:

• Critical points: Where $f'(x) = 0$ or is undefined.

• Second derivative test: $f''(x)$ can help determine if a critical point is a max or min.

Step-by-Step Guidance

1. Use the critical points found in Q2b: $x = 0$ and $x = 2$.

2. Compute $f(x)$ at these points to find the function values.

3. Use the second derivative or sign analysis to determine if each is a maximum or minimum.

4. Sketch the graph based on the behavior at infinity and the local extrema, but do not draw or describe the full graph yet.

Try solving on your own before revealing the answer!

Final Answer:

At $x = 0$, $f(0) = 4$; at $x = 2$, $f(2) = 0$.

$x = 0$ is a local maximum, $x = 2$ is a local minimum.

The graph rises for $x < 0$, falls between $0 < x < 2$, then rises again for $x > 2$.

Q3a. Låt $f(x) = \sqrt{x}$. Bestäm Taylorpolynomet av grad 2 till $f$ kring $x = 25$ och använd det för att bestämma ett närmevärde till $\sqrt{27}$.

Background

Topic: Taylor Series Approximation

This question tests your ability to construct a Taylor polynomial and use it for approximation.

Key Terms and Formulas:

• Taylor polynomial of degree 2: $P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2$.

• For $f(x) = \sqrt{x}$, compute derivatives at $x = 25$.

Step-by-Step Guidance

1. Compute $f(25)$, $f'(25)$, and $f''(25)$.

2. Write the Taylor polynomial: $P_2(x) = f(25) + f'(25)(x-25) + \frac{f''(25)}{2}(x-25)^2$.

3. Plug $x = 27$ into $P_2(x)$ to approximate $\sqrt{27}$.

4. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

$f(25) = 5$, $f'(25) = \frac{1}{2 \cdot 5} = 0.1$, $f''(25) = -\frac{1}{4 \cdot 125} = -0.002$.

$P_2(27) = 5 + 0.1 \cdot 2 + \frac{-0.002}{2} \cdot 4 = 5 + 0.2 - 0.004 = 5.196$

So, $\sqrt{27} \approx 5.196$ using the Taylor polynomial.

Q3b. Kan vi vara säkra på att storleken på felet i approximationen i (a) är mindre än $\frac{1}{5000}$?

Background

Topic: Taylor Remainder/Approximation Error

This question tests your ability to estimate the error in a Taylor approximation.

Key Terms and Formulas:

• Taylor remainder: $R_2(x) = \frac{f'''(c)}{6}(x-a)^3$ for some $c$ between $a$ and $x$.

• For $f(x) = \sqrt{x}$, compute $f'''(x)$.

Step-by-Step Guidance

1. Compute $f'''(x)$ for $f(x) = \sqrt{x}$.

2. Estimate the maximum value of $|f'''(c)|$ for $c$ between $25$and $27$.

3. Calculate $|R_2(27)| = \left| \frac{f'''(c)}{6} (27-25)^3 \right|$.

4. Set up the inequality $|R_2(27)| < \frac{1}{5000}$, but do not solve it yet.

Try solving on your own before revealing the answer!

Final Answer:

$f'''(x) = \frac{3}{8} x^{-5/2}$.

For $x$ between $25$and $27$, $f'''(x)$ is maximized at $x = 25$.

$|R_2(27)| = \frac{3}{8 \cdot 6} \cdot 25^{-5/2} \cdot 8 = \frac{1}{16 \cdot 625} = \frac{1}{10000}$.

So, the error is less than $\frac{1}{5000}$.

Q4a. Beräkna den generaliserade integralen $\int_1^{\infty} \frac{x}{e^x} \, dx$

Background

Topic: Improper Integrals

This question tests your ability to evaluate an improper integral, possibly using integration by parts.

Key Terms and Formulas:

• Improper integral: $\int_a^{\infty} f(x) dx = \lim_{b \to \infty} \int_a^b f(x) dx$.

• Integration by parts: $\int u dv = uv - \int v du$.

Step-by-Step Guidance

1. Rewrite the integrand: $\frac{x}{e^x} = x e^{-x}$.

2. Set up integration by parts: Let $u = x$, $dv = e^{-x} dx$.

3. Compute $du$ and $v$, then apply the integration by parts formula.

4. Set up the definite integral from $x = 1$ to $x = b$, and take the limit as $b \to \infty$, but do not evaluate yet.

Try solving on your own before revealing the answer!

Final Answer:

Integration by parts gives $-x e^{-x} - e^{-x}$ evaluated from $x = 1$ to $x = \infty$.

At $x = \infty$, both terms go to $0$; at$x = 1$,$-1/e - 1/e = -2/e$.

So, the final answer is $2/e$.

Q4b. Avgör om den generaliserade integralen $\int_1^{\infty} \frac{x+1}{2e^x + \cos x} dx$ är konvergent.

Background

Topic: Convergence of Improper Integrals

This question tests your ability to determine whether an improper integral converges, often using comparison tests.

Key Terms and Formulas:

• Comparison test: Compare the integrand to a function with known convergence.

• Asymptotic behavior: For large $x$, $e^x$ dominates $\cos x$.

Step-by-Step Guidance

1. For large $x$, $2e^x + \cos x \approx 2e^x$.

2. So, $\frac{x+1}{2e^x + \cos x} \approx \frac{x+1}{2e^x}$.

3. Compare to $\frac{x}{e^x}$, which is known to converge (see Q4a).

4. Set up the comparison, but do not state the conclusion yet.

Try solving on your own before revealing the answer!

Final Answer:

Since $\frac{x+1}{2e^x + \cos x} < \frac{x+1}{2e^x}$ for large $x$, and $\int_1^{\infty} \frac{x}{e^x} dx$ converges, the given integral also converges by the comparison test.

Q5a. Beräkna volymen av den kropp som uppstår då området som begränsas av astroidkurvan $x^{2/3} + y^{2/3} = 1$ och koordinataxlarna roteras ett varv runt x-axeln.

Background

Topic: Volumes of Revolution (Parametric Curves)

This question tests your ability to compute the volume generated by rotating a region bounded by a parametric curve about the x-axis.

Key Terms and Formulas:

• Volume of revolution: $V = \int_a^b \pi y^2 dx$ or, for parametric curves, $V = \int_{t_1}^{t_2} \pi y^2 \frac{dx}{dt} dt$.

• Parametrization: $x = \cos^3 t$, $y = \sin^3 t$, $0 \leq t \leq \frac{\pi}{2}$.

Step-by-Step Guidance

1. Set up the volume integral using the parametric form: $V = \int_{0}^{\pi/2} \pi (\sin^3 t)^2 \frac{d}{dt}(\cos^3 t) dt$.

2. Compute $y^2 = \sin^6 t$ and $\frac{dx}{dt} = -3 \cos^2 t \sin t$.

3. Substitute into the integral: $V = \int_{0}^{\pi/2} \pi \sin^6 t \cdot (-3 \cos^2 t \sin t) dt$.

4. Set up the integral, but do not evaluate yet.

Try solving on your own before revealing the answer!

Final Answer:

$V = \int_{0}^{\pi/2} \pi \sin^6 t \cdot (-3 \cos^2 t \sin t) dt$

After simplification and integrating, the final answer is $\frac{3\pi}{16}$.

Q5b. Beräkna längden av astroidkurvan $x = \cos^3 t$, $y = \sin^3 t$, $0 \leq t \leq \frac{\pi}{2}$.

Background

Topic: Arc Length of Parametric Curves

This question tests your ability to compute the length of a curve given in parametric form.

Key Terms and Formulas:

• Arc length: $L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$.

• Parametrization: $x = \cos^3 t$, $y = \sin^3 t$.

Step-by-Step Guidance

1. Compute $\frac{dx}{dt} = -3 \cos^2 t \sin t$ and $\frac{dy}{dt} = 3 \sin^2 t \cos t$.

2. Set up the arc length integral: $L = \int_{0}^{\pi/2} \sqrt{(-3 \cos^2 t \sin t)^2 + (3 \sin^2 t \cos t)^2} dt$.

3. Simplify the expression under the square root.

4. Set up the integral, but do not evaluate yet.

Try solving on your own before revealing the answer!

Final Answer:

After simplification, $L = 3 \int_{0}^{\pi/2} \cos t \sin t dt$.

Evaluating, the final answer is $\frac{3\pi}{8}$.

Q6a. Avgör om serien $\sum_{k=3}^{\infty} \frac{1}{k \ln k}$ är konvergent eller divergent.

Background

Topic: Series Convergence (Integral Test)

This question tests your ability to determine convergence of a series using the integral test.

Key Terms and Formulas:

• Integral test: If $f(k)$ is positive, decreasing, and continuous, then $\sum f(k)$ converges iff $\int f(x) dx$ converges.

• Compare to $\sum \frac{1}{k}$, which diverges.

Step-by-Step Guidance

1. Set up the integral $\int_3^{\infty} \frac{1}{x \ln x} dx$.

2. Use substitution: Let $u = \ln x$, $du = \frac{1}{x} dx$.

3. Rewrite the integral in terms of $u$ and analyze its convergence.

4. Set up the comparison, but do not state the conclusion yet.

Try solving on your own before revealing the answer!

Final Answer:

Substitution gives $\int_{\ln 3}^{\infty} \frac{1}{u} du$, which diverges.

Therefore, the series $\sum_{k=3}^{\infty} \frac{1}{k \ln k}$ diverges.

Q6b. Avgör om serien $\sum_{k=1}^{\infty} \frac{1}{k + \ln(k + 1)}$ är konvergent eller divergent.

Background

Topic: Series Convergence (Comparison Test)

This question tests your ability to determine convergence of a series by comparing to a known divergent series.

Key Terms and Formulas:

• Comparison test: Compare to $\sum \frac{1}{k}$.

• For large $k$, $\ln(k+1)$ grows slowly compared to $k$.

Step-by-Step Guidance

1. For large $k$, $k + \ln(k+1) \approx k$, so $\frac{1}{k + \ln(k+1)} \approx \frac{1}{k}$.

2. Since $\sum \frac{1}{k}$ diverges, compare the two series.

3. Set up the comparison, but do not state the conclusion yet.

Try solving on your own before revealing the answer!

Final Answer:

Since $\frac{1}{k + \ln(k+1)} > \frac{1}{2k}$ for large $k$, and $\sum \frac{1}{k}$ diverges, the given series also diverges.