Final, Winter 2024

Find $G'(x)$, given:

$$G(x) = \int_{1}^{x^3} \frac{5+t}{1+\sin^2(t)} dt$$



We need to find the derivative $G'(x)$ of a function defined by a definite integral with a variable upper limit.



Using the Fundamental Theorem of Calculus:

If $G(x) = \int_{a}^{g(x)} f(t) dt$, then $G'(x) = f(g(x)) \cdot g'(x)$

In our case:
- $g(x) = x^3$ (the upper limit of integration)
- $f(t) = \frac{5+t}{1+\sin^2(t)}$ (the integrand)
- $g'(x) = 3x^2$ (the derivative of the upper limit)









Therefore:

$$\boxed{G'(x) = \frac{15x^2+3x^5}{1+\sin^2(x^3)}}$$



Your professor designed this problem to test your understanding of the Fundamental Theorem of Calculus with variable limits.

Find the area of one complete positive region bounded by the curve $y = 2\cos(x)$ and the x-axis.



We need to find the area of one complete positive region bounded by:
- The curve $y = 2\cos(x)$
- The x-axis where $y = 0$



At the x-axis, $y = 0$:

$$0 = 2\cos(x)$$
$$\cos(x) = 0$$

This occurs at $x = \frac{\pi}{2} + n\pi$ for integer values of $n$.

Since we want one complete positive region, we need to identify where the function is positive. The cosine function is positive in the interval $(-\frac{\pi}{2}, \frac{\pi}{2}) + 2n\pi$ for integer values of $n$.

For simplicity, let's choose the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, which contains one complete positive region.



In the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the cosine function is positive, with its maximum value of 1 at $x = 0$. This means our function $y = 2\cos(x)$ is non-negative throughout the interval and reaches a maximum value of 2 at $x = 0$.



For a region bounded by $y = f(x)$, the x-axis, $x = a$, and $x = b$, the area is $\int_{a}^{b} f(x) \, dx$

For our problem:
$$\text{Area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2\cos(x) \, dx$$







Therefore:
$$\boxed{\text{Area} = 4 \text{ square units}}$$

Evaluate $\int \sqrt{16 - x^2} \, dx$.



This is an integral of the form $\int \sqrt{a^2 - x^2} \, dx$, where $a = 4$. This type of integral can be evaluated using trigonometric substitution.



When you have $\sqrt{a^2 - x^2}$, substitute $x = a\sin\theta$ where $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$

Let $x = 4\sin\theta$, which gives us:
$$dx = 4\cos\theta \, d\theta$$

For the radical term:


Note: We can drop the absolute value signs since $\cos\theta \geq 0$ in the interval $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.



Substituting into our integral:




We can use the identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$:





We need to express $\theta$ and $\sin(2\theta)$ in terms of $x$.

Since $x = 4\sin\theta$, we have:
$$\sin\theta = \frac{x}{4}$$
$$\theta = \arcsin\frac{x}{4}$$

For $\sin(2\theta)$, we use the double-angle formula:


Substituting back:


Therefore:





This problem tests your ability to recognize an integral that requires trigonometric substitution

Evaluate $\int_{-x^3}^{x^3} t^3 \sin(t^4) \, dt$.



Always check for symmetry properties first when evaluating definite integrals over intervals like [-a,a]. This can save significant calculation time.

Let's examine if the integrand $t^3 \sin(t^4)$ has symmetry properties:
- $t^4$ is even (same value for $t$ and $-t$)
- $\sin(t^4)$ will have the same value at $t$ and $-t$ since $t^4$ is even
- $t^3$ is odd (changes sign when $t$ changes to $-t$)

Therefore, $t^3 \sin(t^4)$ is an odd function because an even function multiplied by an odd function results in another odd function:
$$(-t)^3 \sin((-t)^4) = -t^3 \sin(t^4)$$



For any odd function $f(-t) = -f(t)$, the definite integral over a symmetric interval $[-a, a]$ is always zero:

$$\int_{-a}^{a} f(t) \, dt = 0$$

Since our integrand is odd and the interval $[-x^3, x^3]$ is symmetric about the origin, we can immediately conclude:

$$\boxed{\int_{-x^3}^{x^3} t^3 \sin(t^4) \, dt = 0}$$



This problem tests your ability to:

1. Recognize and apply symmetry properties of functions
2. Understand how odd functions behave when integrated over symmetric intervals

The professor included this problem to teach you to always look for structural properties and symmetry before attempting more complicated integration techniques. Recognizing oddness/evenness can save significant time and effort in many calculus problems.

Evaluate $\int_{0}^{\pi/2} \sin^5 x \cos^2 x \, dx$.



This is an integral involving powers of sine and cosine. We can approach this using substitution or by rewriting in terms of other trigonometric functions.



For integrals with $\sin^m x \cos^n x$, when at least one exponent is odd, substitute the function with the odd exponent in terms of the other.

Since $\sin^5 x$ has an odd exponent, let's rewrite it as:
$$\sin^5 x$$
$$= \sin^4 x \cdot \sin x = (1-\cos^2 x)^2 \cdot \sin x$$


Let's substitute $u = \cos x$, which gives $du = -\sin x \, dx$ or $\sin x \, dx = -du$.

When $x = 0$, $u = \cos(0) = 1$
When $x = \pi/2$, $u = \cos(\pi/2) = 0$

Rewriting our integral:




Expanding $(1-u^2)^2$:


So our integral becomes:










Therefore:




This problem tests your ability to:

1. Work with trigonometric integrals involving multiple powers
2. Make appropriate substitutions to simplify complex expressions
3. Use algebraic manipulation to expand expressions strategically

This type of integral frequently appears in physics and engineering applications, particularly in electromagnetic theory and signal processing.

Evaluate $\int x^2 \arccos(x^3) \, dx$.



This is an integral containing a product of a polynomial and an inverse trigonometric function. We can use integration by parts to handle it.



When integrating products where one factor is an inverse trigonometric function, choose that function as $u$ and the remaining factors as $dv$.

Let's use the formula:
$$\int u \, dv = uv - \int v \, du$$

Let:
- $u = \arccos(x^3)$
- $dv = x^2 \, dx$

Then:
- $du = \frac{-3x^2}{\sqrt{1-(x^3)^2}} \, dx = \frac{-3x^2}{\sqrt{1-x^6}} \, dx$
- $v = \int x^2 \, dx = \frac{x^3}{3}$







For the integral $\int \frac{x^5}{\sqrt{1-x^6}} \, dx$, we can use the substitution $u = x^6$:



This gives us:






Therefore:




This problem tests your ability to:

1. Handle more complex integration by parts scenarios
2. Recognize the need for multiple substitutions
3. Deal with derivatives of inverse trigonometric functions
4. Manage algebraic complexity while keeping track of all terms

Find the area of the region enclosed between $y = \sin^2 x$ and $y = \cos^2 x$ from $x = 0$ to $x = \frac{\pi}{2}$.



We need to find the area between the curves $y = \sin^2 x$ and $y = \cos^2 x$ over the interval $[0, \frac{\pi}{2}]$.

First, let's evaluate these functions at the endpoints:

At $x = 0$:
$$\sin^2(0) = 0$$
$$\cos^2(0) = 1$$

At $x = \frac{\pi}{2}$:
$$\sin^2(\frac{\pi}{2}) = 1$$
$$\cos^2(\frac{\pi}{2}) = 0$$



Let's find where $\sin^2 x = \cos^2 x$:
$$\sin^2 x = \cos^2 x$$

Since $\sin^2 x + \cos^2 x = 1$, we have:
$$\sin^2 x = \cos^2 x = \frac{1}{2}$$

This occurs when $\sin x = \cos x = \frac{1}{\sqrt{2}}$, which happens at $x = \frac{\pi}{4}$.



Based on our analysis:
- For $0 < x < \frac{\pi}{4}$: $\cos^2 x > \sin^2 x$
- For $\frac{\pi}{4} < x < \frac{\pi}{2}$: $\sin^2 x > \cos^2 x$

When finding area between curves that cross, split the integral at intersection points and use absolute value of the difference in each region.

Using the tip: the area is calculated as:






When working with expressions like $\sin^2 x - \cos^2 x$ or $\cos^2 x - \sin^2 x$, always try to apply the double angle formula $\cos(2x) = \cos^2 x - \sin^2 x$ to simplify integration.


We can simplify using the identity $\cos^2 x - \sin^2 x = \cos(2x)$:











Therefore:




This problem tests your ability to:

1. Recognize trigonometric identities that simplify the calculation
2. Find intersection points of curves algebraically
3. Set up integrals correctly based on where one curve is above the other
4. Use the double-angle formula for cosine to simplify integration

Find the general solution y(t) of the differential equation $\frac{dy}{dt} = t^2e^{-3y}$.



First, I need to identify this as a separable differential equation. The equation can be written with y-terms on one side and t-terms on the other.

For a separable equation $\frac{dy}{dt} = g(t)h(y)$, rearrange to get $\frac{1}{h(y)}dy = g(t)dt$ and then integrate both sides.

Rewriting the equation:
$$\frac{dy}{dt} = t^2e^{-3y}$$

$$e^{3y}dy = t^2dt$$



Let's integrate both sides:

For the left side:
$$\int e^{3y}dy = \frac{1}{3}e^{3y} + C_1$$

For the right side:
$$\int t^2dt = \frac{t^3}{3} + C_2$$

Putting these together with $C = C_2 - C_1$:
$$\frac{1}{3}e^{3y} = \frac{t^3}{3} + C$$



Multiply both sides by 3:
$$e^{3y} = t^3 + 3C$$

Taking the natural logarithm of both sides:
$$3y = \ln(t^3 + 3C)$$

$$y = \frac{1}{3}\ln(t^3 + 3C)$$

We can simplify by setting K = 3C:
$$y = \frac{1}{3}\ln(t^3 + K)$$

This is our general solution, where K is an arbitrary constant.

$$\boxed{y(t) = \frac{1}{3}\ln(t^3 + K)}$$



Your professor assigned this question to test:
1. Your ability to recognize and solve separable differential equations
2. Your comfort with combining integration and logarithm techniques
3. Your understanding of how to handle constants of integration when dealing with nonlinear equations

For the improper integral $$\int_{1}^{\infty} \frac{x^2 + \sin x}{x^3 + x\ln x} dx$$, you need to analyze the behavior of the integrand as x approaches infinity.

First, let's look at the asymptotic behavior by focusing on the highest-degree terms:

$$\frac{x^2 + \sin x}{x^3 + x\ln x} \sim \frac{x^2}{x^3} = \frac{1}{x} \text{ as } x \to \infty$$

When analyzing improper integrals with limits at infinity, compare the integrand to well-known convergent or divergent functions like $\frac{1}{x^p}$.



Since our integrand behaves like $\frac{1}{x}$ for large values of x, and we know that $$\int_{1}^{\infty} \frac{1}{x} \, dx$$ diverges (it’s the p-integral with $p = 1$), we use the Comparison Test. We aim to find a positive constant $k$ such that for sufficiently large $x$:

$$\frac{x^2 + \sin x}{x^3 + x \ln x} > \frac{k}{x},$$

where $\int_{1}^{\infty} \frac{k}{x} \, dx$ diverges, implying our integral diverges if the inequality holds.



Since $-1 \leq \sin x \leq 1$ for all $x$, we have:
$$x^2 - 1 \leq x^2 + \sin x \leq x^2 + 1.$$

For $x > 2$, $x^2 - 1 > \frac{x^2}{2}$, because:
$$x^2 - 1 > \frac{x^2}{2}$$
$$\implies x^2 - \frac{x^2}{2} > 1$$
$$\implies \frac{x^2}{2} > 1$$
$$\implies x^2 > 2$$
$$\implies x > \sqrt{2} \approx 1.414$$

and $x > 2$ satisfies this.

Thus:
$$x^2 + \sin x > \frac{x^2}{2}.$$

For the denominator, since $\ln x$ grows slower than $x$, for sufficiently large $x$ (e.g., $x > 10$):
$$x^3 + x \ln x < x^3 + x^3 = 2x^3,$$


Combining these:
$$\frac{x^2 + \sin x}{x^3 + x \ln x} > \frac{x^2 / 2}{2x^3} = \frac{x^2}{4x^3} = \frac{1}{4x}.$$

So our lower bound is $\frac{1}{4x}$.

We gave the rigorous work here, but for only 3 marks you can just state $x^2 + \sin x > \frac{x^2}{2}$ and $x^3 + x \ln x < x^3 + x^3 = 2x^3$ without showing any work. But you SHOULD have an intuitive understanding that both of these inequalities are true!



Since $$\frac{x^2 + \sin x}{x^3 + x\ln x} > \frac{1}{4x}$$ for sufficiently large x, and we know that $$\int_{1}^{\infty} \frac{1}{4x} dx = \frac{1}{4}\int_{1}^{\infty} \frac{1}{x} dx$$ diverges, by the Comparison Test, our original integral also diverges.

$$\boxed{\text{The improper integral diverges}}$$



Your professor assigned this question to test:
1. Your understanding of how to analyze limiting behavior of complex rational functions
2. Your ability to apply comparison tests for improper integrals
3. Your knowledge of which standard improper integrals (like $\int \frac{1}{x^p} dx$) converge or diverge




Consider the improper integral:
$$\int_{1}^{\infty} \frac{x^2 + \sin x}{x^3 + x \ln x} \, dx.$$

To determine convergence, analyze the integrand as $x \to \infty$. The dominant terms suggest:
$$\frac{x^2 + \sin x}{x^3 + x \ln x} \sim \frac{x^2}{x^3} = \frac{1}{x},$$
since $\sin x$ is bounded and $x \ln x < x^2$ for large $x$.

We know:
$$\int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{b \to \infty} \ln x \big|_{1}^{b}$$
$$= \lim_{b \to \infty} (\ln b - \ln 1)$$
$$= \lim_{b \to \infty} (\ln b - 0)$$
$$= \lim_{b \to \infty} \ln b$$
$$= \infty$$

which diverges.



Let:
- $f(x) = \frac{x^2 + \sin x}{x^3 + x \ln x}$,
- $g(x) = \frac{1}{x}$.

Both are positive for $x > 1$. Compute the limit:
$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{\frac{x^2 + \sin x}{x^3 + x \ln x}}{\frac{1}{x}}$$
$$= \lim_{x \to \infty} \frac{x^2 + \sin x}{x^3 + x \ln x} \cdot x$$
$$= \lim_{x \to \infty} \frac{x^3 + x \sin x}{x^3 + x \ln x}$$

Factorize:
$$= \lim_{x \to \infty} \frac{x^3 (1 + \frac{\sin x}{x^2})}{x^3 (1 + \frac{\ln x}{x^2})}$$
$$= \lim_{x \to \infty} \frac{1 + \frac{\sin x}{x^2}}{1 + \frac{\ln x}{x^2}}$$

Evaluate:
- $\frac{\sin x}{x} \to 0$ as $x \to \infty$ (since $|\sin x| \leq 1$),
- $\frac{\ln x}{x} \to 0$ as $x \to \infty$ (since $\ln x$ grows slower than $x$).

Thus:
$$\lim_{x \to \infty} \frac{1 + \frac{\sin x}{x}}{1 + \frac{\ln x}{x}} = \frac{1 + 0}{1 + 0} = 1.$$

The limit is a positive finite constant.



By the Limit Comparison Test: if $\lim_{x \to \infty} \frac{f(x)}{g(x)} = c$ where $0 < c < \infty$, and $\int_{1}^{\infty} g(x) \, dx$ diverges, then $\int_{1}^{\infty} f(x) \, dx$ diverges. Since $\int_{1}^{\infty} \frac{1}{x} \, dx$ diverges and the limit is 1, the original integral diverges.

$$\boxed{\text{The improper integral diverges}}$$



This approach tests:
1. Recognition of asymptotic behavior in rational functions with oscillatory terms,
2. Mastery of the Limit Comparison Test for improper integrals,
3. Familiarity with standard integrals like $\int_{1}^{\infty} \frac{1}{x} \, dx$.

For the integral
$$\int_{0}^{2} \frac{dx}{(2-x)^{4/3}}$$
you need to identify where the integrand becomes improper.

The denominator
$$(2-x)^{4/3}$$
equals zero when $x = 2$, which is the upper limit of integration. This makes the integral improper since the integrand becomes unbounded at $x = 2$.

When evaluating improper integrals, check if the integrand becomes unbounded at any point in the interval of integration, including the endpoints.



To handle the improper behavior at $x = 2$, rewrite the integral as a limit:

$$\int_{0}^{2} \frac{dx}{(2-x)^{4/3}}$$
$$= \lim_{b \to 2^-} \int_{0}^{b} \frac{dx}{(2-x)^{4/3}}$$



First, find the indefinite integral:

$$\int \frac{dx}{(2-x)^{4/3}}$$

Use substitution with $u = 2-x$, so $du = -dx$ or $dx = -du$:

$$\int \frac{dx}{(2-x)^{4/3}}$$
$$= -\int \frac{du}{u^{4/3}}$$
$$= -\int u^{-4/3} \, du$$

Apply the power rule:

$$-\int u^{-4/3} \, du$$
$$= -\frac{u^{-4/3+1}}{-4/3+1} + C$$
$$= -\frac{u^{-1/3}}{-1/3} + C$$
$$= 3u^{-1/3} + C$$
$$= \frac{3}{(2-x)^{1/3}} + C$$



Now evaluate using the limits:

$$\lim_{b \to 2^-} \int_{0}^{b} \frac{dx}{(2-x)^{4/3}}$$
$$= \lim_{b \to 2^-} \left[ \frac{3}{(2-x)^{1/3}} \right]_{0}^{b}$$
$$= \lim_{b \to 2^-} \left( \frac{3}{(2-b)^{1/3}} - \frac{3}{(2-0)^{1/3}} \right)$$
$$= \lim_{b \to 2^-} \left( \frac{3}{(2-b)^{1/3}} - \frac{3}{2^{1/3}} \right)$$

As $b$ approaches $2$, the term
$$\frac{3}{(2-b)^{1/3}}$$
approaches infinity:

$$\lim_{b \to 2^-} \frac{3}{(2-b)^{1/3}} = \infty$$

Therefore:

$$\lim_{b \to 2^-} \left( \frac{3}{(2-b)^{1/3}} - \frac{3}{2^{1/3}} \right)$$
$$= \infty - \frac{3}{2^{1/3}}$$
$$= \infty$$



Since the limit diverges to infinity, the improper integral diverges:

$$\boxed{\text{The improper integral } \int_{0}^{2} \frac{dx}{(2-x)^{4/3}} \text{ diverges}}$$