Step-by-Step Calculus Guidance: Integration Techniques & Differential Equations

Q1. Evaluate $\int \frac{2x + 5}{x^2 + 4x + 13} \, dx$.

Background

Topic: Basic Antiderivatives & Substitution

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

Key Terms and Formulas:

• Substitution: Useful when the numerator is related to the derivative of the denominator.

• Standard forms: $\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$

Step-by-Step Guidance

1. Notice that the denominator $x^2 + 4x + 13$ can be rewritten by completing the square: $x^2 + 4x + 13 = (x + 2)^2 + 9$.

2. Check if the numerator $2x + 5$ can be expressed in terms of the derivative of the denominator. The derivative of $x^2 + 4x + 13$ is $2x + 4$.

3. Rewrite $2x + 5$ as $(2x + 4) + 1$ to split the integral into two parts.

4. Set up the integral as $\int \frac{2x + 4}{x^2 + 4x + 13} \, dx + \int \frac{1}{x^2 + 4x + 13} \, dx$.

5. For the first part, use substitution $u = x^2 + 4x + 13$, $du = 2x + 4 \, dx$. For the second part, use the completed square form and the standard arctangent formula.

Try solving on your own before revealing the answer!

Final Answer:

$\int \frac{2x + 5}{x^2 + 4x + 13} \, dx = \ln|x^2 + 4x + 13| + \frac{1}{3} \arctan\left(\frac{x + 2}{3}\right) + C$

The integral was split using algebraic manipulation, and each part was integrated using substitution and the arctangent formula.

Q2. Evaluate $\int x^2 e^{2x} \, dx$.

Background

Topic: Integration by Parts

This question tests your ability to use integration by parts, especially for products of polynomials and exponentials.

Key Terms and Formulas:

• Integration by Parts: $\int u \, dv = u v - \int v \, du$

• Tabular integration: Useful for repeated integration by parts.

Step-by-Step Guidance

1. Choose $u = x^2$ (polynomial) and $dv = e^{2x} \, dx$ (exponential).

2. Compute $du = 2x \, dx$ and $v = \frac{1}{2} e^{2x}$.

3. Apply integration by parts: $\int x^2 e^{2x} \, dx = x^2 \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \cdot 2x \, dx$.

4. Simplify the integral: $\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \int x e^{2x} \, dx$.

5. Set up integration by parts again for $\int x e^{2x} \, dx$ (let $u = x$, $dv = e^{2x} \, dx$).

Try solving on your own before revealing the answer!

Final Answer:

$\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \frac{1}{2} x e^{2x} + \frac{1}{4} e^{2x} + C$

Repeated integration by parts was used, reducing the polynomial degree each time.

Q3. Evaluate each integral:

(a) $\int \sin^3 x \cos^4 x \, dx$

Background

Topic: Trigonometric Integrals

This question tests your ability to integrate powers of sine and cosine, using reduction formulas or substitution.

Key Terms and Formulas:

• Power-reduction identities: $\sin^2 x = 1 - \cos^2 x$, $\cos^2 x = 1 - \sin^2 x$

• Substitution: Useful when one power is odd.

Step-by-Step Guidance

1. Since the power of sine is odd, factor $\sin^3 x$ as $\sin x \cdot \sin^2 x$.

2. Rewrite $\sin^2 x$ as $1 - \cos^2 x$.

3. Let $u = \cos x$, $du = -\sin x \, dx$.

4. Express the integral in terms of $u$ and set up the new integral.

Try solving on your own before revealing the answer!

Final Answer (a):

$\int \sin^3 x \cos^4 x \, dx = -\frac{1}{5} \cos^5 x + \frac{1}{7} \cos^7 x + C$

Substitution and power-reduction were used to express the integral in terms of cosine.

(b) $\int \tan^3 x \sec^4 x \, dx$

Background

Topic: Trigonometric Integrals

This question tests your ability to integrate products of tangent and secant powers, using substitution or reduction formulas.

Key Terms and Formulas:

• $\frac{d}{dx} \tan x = \sec^2 x$

• $\frac{d}{dx} \sec x = \sec x \tan x$

• Reduction formulas for powers of tangent and secant.

Step-by-Step Guidance

1. Rewrite $\tan^3 x \sec^4 x$ as $\tan^2 x \tan x \sec^4 x$.

2. Recall $\tan^2 x = \sec^2 x - 1$.

3. Express the integral as $\int (\sec^2 x - 1) \tan x \sec^4 x \, dx$.

4. Let $u = \tan x$, $du = \sec^2 x \, dx$.

5. Set up the integral in terms of $u$ and simplify.

Try solving on your own before revealing the answer!

Final Answer (b):

$\int \tan^3 x \sec^4 x \, dx = \frac{1}{4} \tan^4 x + \frac{1}{2} \tan^2 x + C$

Substitution and reduction formulas were used to integrate the expression.

Q4. Evaluate $\int \frac{\sqrt{x^2 - 16}}{x} \, dx$, $x > 4$. Use a trigonometric substitution and include the right triangle used for back-substitution.

Background

Topic: Trigonometric Substitution

This question tests your ability to use trigonometric substitution for integrals involving $\sqrt{x^2 - a^2}$.

Key Terms and Formulas:

• Trigonometric substitution: $x = a \sec \theta$ for $\sqrt{x^2 - a^2}$

• Right triangle for back-substitution: $x = 4 \sec \theta$, $\sqrt{x^2 - 16} = 4 \tan \theta$

Step-by-Step Guidance

1. Let $x = 4 \sec \theta$, so $dx = 4 \sec \theta \tan \theta \, d\theta$.

2. Substitute into the integral: $\sqrt{x^2 - 16} = 4 \tan \theta$, $x = 4 \sec \theta$.

3. Rewrite the integral in terms of $\theta$ and simplify.

4. Integrate with respect to $\theta$.

5. Draw the right triangle with hypotenuse $x$, adjacent $4$, and opposite$\sqrt{x^2 - 16}$ for back-substitution.

Try solving on your own before revealing the answer!

Final Answer:

$\int \frac{\sqrt{x^2 - 16}}{x} \, dx = \sqrt{x^2 - 16} - 4 \ln\left| x + \sqrt{x^2 - 16} \right| + C$

Trigonometric substitution was used, and the right triangle allowed for back-substitution to $x$.

Q5. Evaluate $\int \frac{3x^2 + 8x + 2}{x(x + 1)^2} \, dx$.

Background

Topic: Partial Fractions

This question tests your ability to decompose a rational function into partial fractions and integrate each term.

Key Terms and Formulas:

• Partial fraction decomposition: Express $\frac{3x^2 + 8x + 2}{x(x + 1)^2}$ as $\frac{A}{x} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}$

• Integrate each term separately.

Step-by-Step Guidance

1. Set up the partial fraction decomposition: $\frac{3x^2 + 8x + 2}{x(x + 1)^2} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}$.

2. Multiply both sides by $x(x + 1)^2$ and solve for $A$, $B$, and $C$ by equating coefficients.

3. Once $A$, $B$, and $C$ are found, write the integral as the sum of three simpler integrals.

4. Integrate each term: $\int \frac{A}{x} \, dx$, $\int \frac{B}{x + 1} \, dx$, $\int \frac{C}{(x + 1)^2} \, dx$.

Try solving on your own before revealing the answer!

Final Answer:

$\int \frac{3x^2 + 8x + 2}{x(x + 1)^2} \, dx = 3 \ln|x| + (3 - 1) \ln|x + 1| - \frac{1}{x + 1} + C$

Partial fraction decomposition allowed the integral to be split and solved term by term.

Q6. (a) For each integral below, state the first technique you would try. Do not evaluate these four integrals.

• (i) $\int x \ln x \, dx$

• (ii) $\int \frac{dx}{x^2 - 5x + 6}$

• (iii) $\int \sqrt{9 - x^2} \, dx$

• (iv) $\int \sin^4 x \, dx$

Background

Topic: Choosing Integration Techniques

This question tests your ability to recognize which integration method is most appropriate for different types of integrals.

Key Terms and Formulas:

• Integration by parts: Useful for products like $x \ln x$.

• Partial fractions: Useful for rational functions with factorable denominators.

• Trigonometric substitution: Useful for $\sqrt{a^2 - x^2}$.

• Power-reduction: Useful for even powers of sine or cosine.

Step-by-Step Guidance

1. For (i), recognize the product of a polynomial and a logarithm; integration by parts is likely best.

2. For (ii), factor the denominator and use partial fractions.

3. For (iii), use trigonometric substitution with $x = 3 \sin \theta$.

4. For (iv), use power-reduction identities for $\sin^4 x$.

Try identifying the techniques on your own before revealing the answer!

Final Answer:

• (i) Integration by parts

• (ii) Partial fractions

• (iii) Trigonometric substitution

• (iv) Power-reduction formula

Each integral is matched with the most efficient technique based on its structure.

Q6 (b). Use the supplied table entry $\int \frac{du}{(u^2 + a^2)^{3/2}} = \frac{u}{a^2 \sqrt{u^2 + a^2}} + C$ to evaluate $\int \frac{e^x}{(9 + e^{2x})^{3/2}} \, dx$.

Background

Topic: Table Integration & Substitution

This question tests your ability to use a table formula after an appropriate substitution.

Key Terms and Formulas:

• Substitution: Let $u = e^x$.

• Table formula: $\int \frac{du}{(u^2 + a^2)^{3/2}} = \frac{u}{a^2 \sqrt{u^2 + a^2}} + C$

Step-by-Step Guidance

1. Let $u = e^x$, so $du = e^x \, dx$.

2. Rewrite the integral in terms of $u$: $\int \frac{du}{(9 + u^2)^{3/2}}$.

3. Apply the table formula with $a^2 = 9$, so $a = 3$.

4. Write the result in terms of $u$ and then back-substitute $u = e^x$.

Try solving on your own before revealing the answer!

Final Answer:

$\int \frac{e^x}{(9 + e^{2x})^{3/2}} \, dx = \frac{e^x}{9 \sqrt{9 + e^{2x}}} + C$

Substitution allowed the use of the table formula, and the answer was back-substituted to $x$.

Q7. Evaluate, or state that the integral diverges: $\int_1^{\infty} \frac{\ln x}{x^2} \, dx$.

Background

Topic: Improper Integrals

This question tests your ability to evaluate improper integrals on infinite intervals and determine convergence.

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: Useful for $\ln x$ terms.

Step-by-Step Guidance

1. Set up the improper integral as a limit: $\lim_{b \to \infty} \int_1^b \frac{\ln x}{x^2} \, dx$.

2. Use integration by parts: Let $u = \ln x$, $dv = x^{-2} \, dx$.

3. Compute $du = \frac{1}{x} \, dx$, $v = -x^{-1}$.

4. Apply integration by parts and set up the resulting expression.

5. Evaluate the limit as $b \to \infty$ and check for convergence.

Try solving on your own before revealing the answer!

Final Answer:

$\int_1^{\infty} \frac{\ln x}{x^2} \, dx = 1$

The integral converges, and integration by parts yields a finite value.

Q8. (a) Evaluate, or state that the integral diverges: $\int_0^1 x^{-2/3} \, dx$.

Background

Topic: Improper Integrals & p-Integrals

This question tests your ability to evaluate improper integrals with singularities at endpoints.

Key Terms and Formulas:

• p-integral: $\int_0^1 x^p \, dx$ converges if $p > -1$.

• Set up as a limit if the integrand is unbounded at $x = 0$.

Step-by-Step Guidance

1. Recognize that $x^{-2/3}$ is unbounded at $x = 0$.

2. Set up the integral as $\lim_{a \to 0^+} \int_a^1 x^{-2/3} \, dx$.

3. Integrate $x^{-2/3}$: $\int x^{-2/3} \, dx = \frac{3}{1} x^{1/3}$.

4. Evaluate the limit as $a \to 0^+$.

Try solving on your own before revealing the answer!

Final Answer (a):

$\int_0^1 x^{-2/3} \, dx = 3$

The integral converges because the exponent $-2/3 > -1$.

Q8. (b) Determine whether the integral converges or diverges. Justify with a comparison: $\int_1^{\infty} \frac{dx}{\sqrt{x^3 + 1}}$.

Background

Topic: Improper Integrals & Comparison Test

This question tests your ability to use comparison arguments to determine convergence of improper integrals.

Key Terms and Formulas:

• Comparison test: Compare to a simpler function whose convergence is known.

• For large $x$, $\sqrt{x^3 + 1} \approx x^{3/2}$.

• $\int_1^{\infty} x^{-3/2} \, dx$ converges if exponent $<-1$.

Step-by-Step Guidance

1. For large $x$, $\sqrt{x^3 + 1} \geq x^{3/2}$, so $\frac{1}{\sqrt{x^3 + 1}} \leq x^{-3/2}$.

2. Compare $\int_1^{\infty} \frac{dx}{\sqrt{x^3 + 1}}$ to $\int_1^{\infty} x^{-3/2} \, dx$.

3. Check if the comparison integral converges.

4. Conclude about the original integral based on the comparison.

Try reasoning through the comparison before revealing the answer!

Final Answer (b):

The integral converges. Since $x^{-3/2}$ converges on $[1, \infty)$, and $\frac{1}{\sqrt{x^3 + 1}} \leq x^{-3/2}$ for large $x$, the original integral converges by the comparison test.

Q9. Consider the initial value problem $\frac{dy}{dt} = t(1 + y^2)$, $y(0) = 0$.

• (a) Explain why the equation is separable.

• (b) Solve the initial value problem explicitly for $y(t)$.

• (c) State the maximal open interval containing $t = 0$ on which your solution is defined.

Background

Topic: Separable Differential Equations

This question tests your ability to recognize and solve separable differential equations, and determine the domain of the solution.

Key Terms and Formulas:

• Separable equation: Can be written as $\frac{dy}{dt} = f(t) g(y)$.

• Integrate both sides: $\int \frac{1}{1 + y^2} \, dy = \int t \, dt$.

• Initial value: $y(0) = 0$.

Step-by-Step Guidance

1. Rewrite the equation: $\frac{dy}{dt} = t(1 + y^2)$.

2. Divide both sides by $1 + y^2$ to get $\frac{1}{1 + y^2} \frac{dy}{dt} = t$.

3. Write in differential form: $\frac{1}{1 + y^2} \, dy = t \, dt$.

4. Integrate both sides: $\int \frac{1}{1 + y^2} \, dy = \int t \, dt$.

5. Apply the initial condition to solve for the constant.

Try solving on your own before revealing the answer!

Final Answer:

• (a) The equation is separable because it can be written as $\frac{dy}{dt} = t(1 + y^2)$, allowing variables to be separated.

• (b) $y(t) = \tan\left(\frac{1}{2} t^2\right)$

• (c) The maximal open interval containing $t = 0$ is $\left(-\sqrt{\frac{\pi}{2}}, \sqrt{\frac{\pi}{2}}\right)$, where $\tan\left(\frac{1}{2} t^2\right)$ is defined.

Separation of variables and integration led to the explicit solution, and the domain is determined by the tangent function.

Q10. A population $P(t)$ satisfies $\frac{dP}{dt} = 0.2P \left(1 - \frac{P}{500}\right)$, $P(0) = 50$.

• (a) Solve the initial value problem.

• (b) Identify the carrying capacity.

• (c) Find the exact time at which the population reaches 250.

Background

Topic: Logistic Differential Equations

This question tests your ability to solve logistic equations, identify carrying capacity, and solve for specific values.

Key Terms and Formulas:

• Logistic equation: $\frac{dP}{dt} = rP \left(1 - \frac{P}{K}\right)$

• Carrying capacity $K$ is the value where population stabilizes.

• Solution: $P(t) = \frac{K}{1 + A e^{-r t}}$

Step-by-Step Guidance

1. Identify $r = 0.2$, $K = 500$, $P(0) = 50$.

2. Write the logistic solution form: $P(t) = \frac{500}{1 + A e^{-0.2 t}}$.

3. Use the initial condition to solve for $A$.

4. Set $P(t) = 250$ and solve for $t$.

Try solving on your own before revealing the answer!

Final Answer:

• (a) $P(t) = \frac{500}{1 + 9 e^{-0.2 t}}$

• (b) Carrying capacity is $500$.

• (c) Population reaches $250$at$t = \frac{1}{0.2} \ln 9 = 11.0$ (rounded to one decimal place).

The logistic model was solved using the standard formula, and the carrying capacity is the limiting value.

Q11. A cup of liquid is initially at $90^\circ$C and is placed in a room at $20^\circ$C. After 5 minutes, the liquid has cooled to $60^\circ$C. Assume Newton's Law of Cooling: $\frac{dT}{dt} = -k(T - 20)$, $k > 0$.

• (a) Solve for $T(t)$, including the exact value of $k$.

• (b) Find the exact time at which the liquid reaches $30^\circ$C, and then give a decimal approximation to the nearest tenth of a minute.

• (c) Identify the equilibrium temperature and state whether it is stable or unstable.

Background

Topic: First-Order Linear Differential Equations & Newton's Law of Cooling

This question tests your ability to solve a linear differential equation, apply initial conditions, and interpret equilibrium.

Key Terms and Formulas:

• Newton's Law of Cooling: $\frac{dT}{dt} = -k(T - T_{room})$

• General solution: $T(t) = T_{room} + (T_0 - T_{room}) e^{-k t}$

• Equilibrium temperature is $T_{room}$.

Step-by-Step Guidance

1. Write the solution: $T(t) = 20 + (90 - 20) e^{-k t}$.

2. Use $T(5) = 60$ to solve for $k$.

3. Set $T(t) = 30$ and solve for $t$.

4. Identify equilibrium temperature and discuss stability.

Try solving on your own before revealing the answer!

Final Answer:

• (a) $T(t) = 20 + 70 e^{-k t}$, $k = \frac{1}{5} \ln \frac{70}{40} \approx 0.112$

• (b) $t = \frac{1}{k} \ln \frac{70}{10} \approx 19.7$ minutes

• (c) Equilibrium temperature is $20^\circ$C, and it is stable.

The solution uses the exponential decay formula, and equilibrium is stable because temperature approaches $20^\circ$C as $t \to \infty$.