Practice Final #4

The integral is $\int_{-1}^{1} (5x^3 - 2x + \arctan x) \, dx$.
The interval of integration is $[-1, 1]$. This is a symmetric interval because it is of the form $[-a, a]$ with $a=1$.
The integrand is $f(x) = 5x^3 - 2x + \arctan x$.


When integrating over a symmetric interval $[-a, a]$, it's useful to check if the integrand $f(x)$ is an odd function or an even function.
- Recall: A function $f$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain.
- Recall: A function $f$ is even if $f(-x) = f(x)$ for all $x$ in its domain.

Let's check the symmetry of $f(x) = 5x^3 - 2x + \arctan x$ by replacing $x$ with $-x$:
$$f(-x)=5(-x{)}^3-2(-x)+\arctan (-x)$$
f(-x) = 5(-x)^3 - 2(-x) + \arctan(-x)

We know that:
- $(-x)^3 = -x^3$
- $\arctan(-x) = -\arctan(x)$ (Arctangent is an odd function)
Substitute these back:
$$\begin{align*}$$
f(-x) &= 5(-x^3) - 2(-x) + (-\arctan x) \\
&= -5x^3 + 2x - \arctan x
\end{align*}
Now, compare this to $-f(x)$:
$$\begin{align*}$$
-f(x) &= -(5x^3 - 2x + \arctan x) \\
&= -5x^3 + 2x - \arctan x
\end{align*}
Since $f(-x) = -f(x)$, the integrand $f(x) = 5x^3 - 2x + \arctan x$ is an odd function.
Alternatively, you could note that $5x^3$ is odd, $-2x$ is odd, and $\arctan x$ is odd. The sum of odd functions is always an odd function.


There is a property for definite integrals over symmetric intervals:
If $f(x)$ is an odd function and continuous on $[-a, a]$, then $\int_{-a}^{a} f(x) \, dx = 0$.
Geometrically, this is because the signed area from $-a$ to 0 cancels out the signed area from 0 to $a$.

Since our integrand $f(x) = 5x^3 - 2x + \arctan x$ is an odd function and the interval is $[-1, 1]$, we can directly apply this property.


$$\int_{-1}^{1} (5x^3 - 2x + \arctan x) \, dx = 0$$
The value of the definite integral is 0. This corresponds to option (A).

$$\boxed{{\displaystyle 0}}$$
\boxed{ 0 }

We need to find $f'(x) = \frac{d}{dx} \left( \int_{h(x)}^{g(x)} F(t) \, dt \right)$.
The Fundamental Theorem of Calculus, Part 1 (FTC Part 1), extended for variable limits of integration, states:

$$\frac{d}{dx}({\int }_{h(x)}^{g(x)}F(t)\hspace{0.17em}dt)=F(g(x))\cdot g^{\mathrm{\prime }}(x)-F(h(x))\cdot h^{\mathrm{\prime }}(x)$$

\frac{d}{dx} \left( \int_{h(x)}^{g(x)} F(t) \, dt \right) = F(g(x)) \cdot g'(x) - F(h(x)) \cdot h'(x)




This formula combines FTC Part 1 ($\frac{d}{dx} \int_a^x F(t) dt = F(x)$) with the Chain Rule. You evaluate the integrand at the upper limit and multiply by the upper limit's derivative, then subtract the integrand evaluated at the lower limit multiplied by the lower limit's derivative.


In our problem $f(x) = \int_{x^2}^{x^3} \cos(t^2) \, dt$:
- The integrand is $F(t) = \cos(t^2)$.
- The upper limit of integration is $g(x) = x^3$.
- The lower limit of integration is $h(x) = x^2$.


We need the derivatives of the upper and lower limits with respect to $x$:
$$g'(x) = \frac{d}{dx}(x^3) = 3x^2$$
$$h'(x) = \frac{d}{dx}(x^2) = 2x$$


We need to evaluate the integrand $F(t) = \cos(t^2)$ at the upper limit $g(x) = x^3$ and the lower limit $h(x) = x^2$:
$$F(g(x)) = F(x^3) = \cos((x^3)^2) = \cos(x^6)$$
$$F(h(x)) = F(x^2) = \cos((x^2)^2) = \cos(x^4)$$


Now substitute all the pieces into the formula: $f'(x) = F(g(x)) g'(x) - F(h(x)) h'(x)$
$$\begin{align*} f'(x) &= (\cos(x^6)) \cdot (3x^2) - (\cos(x^4)) \cdot (2x) \\ &= 3x^2 \cos(x^6) - 2x \cos(x^4) \end{align*}$$


The derivative is $3x^2 \cos(x^6) - 2x \cos(x^4)$.
Comparing this to the options:
(A) $\cos(x^6) - \cos(x^4)$
(B) $3x^2 \cos(x^6) - 2x \cos(x^4)$
(C) $\sin(x^6)(3x^2) - \sin(x^4)(2x)$
(D) $3x^2 \sin(x^6) - 2x \sin(x^4)$

The correct option is (B).
$$\boxed{{\displaystyle (B)\hspace{0.17em}3x^2\cos (x^6)-2x\cos (x^4)}}$$
\boxed{ (B) \, 3x^2 \cos(x^6) - 2x \cos(x^4) }

Integral: $I_A = \int_{1}^{\infty} \frac{\ln x}{x} dx$
Type: This is an improper integral of Type 1 due to the infinite upper limit.
Method: We can try to evaluate it using a limit, possibly with u-substitution.
Let $u = \ln x$. Then $du = \frac{1}{x} dx$.
Change limits:
- When $x=1$, $u = \ln 1 = 0$.
- As $x \to \infty$, $u = \ln x \to \infty$.
Substitute:
$$\begin{align*} I_A &= \int_{0}^{\infty} u \, du \\ &= \lim_{b\to\infty} \int_{0}^{b} u \, du \\ &= \lim_{b\to\infty} \left[ \frac{u^2}{2} \right]_{0}^{b} \\ &= \lim_{b\to\infty} \left( \frac{b^2}{2} - \frac{0^2}{2} \right) \\ &= \lim_{b\to\infty} \frac{b^2}{2} \\ &= \infty \end{align*}$$
Conclusion: The integral $\int_{1}^{\infty} \frac{\ln x}{x} dx$ diverges.


Integral: $I_B = \int_{0}^{1} \frac{1}{\sqrt[3]{x}} dx$
Type: This is an improper integral of Type 2 because the integrand $\frac{1}{x^{1/3}}$ has an infinite discontinuity (vertical asymptote) at the lower limit $x=0$.
Method: This fits the form of a p-integral at a finite endpoint: $\int_{0}^{a} \frac{1}{x^p} dx$.
Rewrite the integrand: $\frac{1}{\sqrt[3]{x}} = \frac{1}{x^{1/3}}$.
Here, $p = 1/3$.
Recall the convergence rule for Type 2 p-integrals: $\int_{0}^{a} \frac{1}{x^p} dx$ converges if $p < 1$ and diverges if $p \ge 1$.
Since $p = 1/3 < 1$, the integral converges.
Conclusion: The integral $\int_{0}^{1} \frac{1}{\sqrt[3]{x}} dx$ converges.


Integral: $I_C = \int_{1}^{\infty} \frac{x}{x^2+1} dx$
Type: This is an improper integral of Type 1 due to the infinite upper limit.
Method 1: Evaluation using u-substitution.
Let $u = x^2+1$. Then $du = 2x \, dx$, so $x \, dx = \frac{1}{2} du$.
Change limits:
- When $x=1$, $u = 1^2+1 = 2$.
- As $x \to \infty$, $u = x^2+1 \to \infty$.
Substitute:
$$\begin{align*} I_C &= \int_{2}^{\infty} \frac{1}{u} \left( \frac{1}{2} du \right) \\ &= \frac{1}{2} \int_{2}^{\infty} \frac{1}{u} \, du \\ &= \frac{1}{2} \lim_{b\to\infty} \int_{2}^{b} \frac{1}{u} \, du \\ &= \frac{1}{2} \lim_{b\to\infty} [\ln|u|]_{2}^{b} \\ &= \frac{1}{2} \lim_{b\to\infty} (\ln b - \ln 2) \\ &= \frac{1}{2} (\infty - \ln 2) \\ &= \infty \end{align*}$$
Conclusion (Method 1): The integral $\int_{1}^{\infty} \frac{x}{x^2+1} dx$ diverges.
Method 2: Limit Comparison Test (LCT).
Compare $a_n = \frac{x}{x^2+1}$ with $b_n = \frac{x}{x^2} = \frac{1}{x}$.
$$\begin{align*} \lim_{x\to\infty} \frac{a_n}{b_n} &= \lim_{x\to\infty} \frac{x/(x^2+1)}{1/x} \\ &= \lim_{x\to\infty} \frac{x^2}{x^2+1} \\ &= \lim_{x\to\infty} \frac{1}{1+1/x^2} \\ &= \frac{1}{1+0} \\ &= 1 \end{align*}$$
Since the limit is finite and positive (1), and $\int_{1}^{\infty} \frac{1}{x} dx$ diverges (p-integral with $p=1$), $I_C$ also diverges by LCT.


Integral: $I_D = \int_{0}^{\pi/2} \tan x \, dx$
Type: This is an improper integral of Type 2 because $\tan x = \frac{\sin x}{\cos x}$ has an infinite discontinuity (vertical asymptote) at the upper limit $x=\pi/2$ (where $\cos x = 0$).
Method: Evaluate using the limit definition.
The antiderivative of $\tan x$ is $-\ln|\cos x|$.
$$\begin{align*} I_D &= \lim_{b\to (\pi/2)^{-}} \int_{0}^{b} \tan x \, dx \\ &= \lim_{b\to (\pi/2)^{-}} [-\ln|\cos x|]_{0}^{b} \\ &= \lim_{b\to (\pi/2)^{-}} (-\ln|\cos b|) - (-\ln|\cos 0|) \\ &= \lim_{b\to (\pi/2)^{-}} (-\ln|\cos b|) + \ln|1| \\ &= \lim_{b\to (\pi/2)^{-}} (-\ln|\cos b|) \end{align*}$$
As $b \to (\pi/2)^{-}$, $\cos b \to 0^{+}$. Therefore, $|\cos b| \to 0^{+}$.
As the argument of $\ln$ approaches $0^{+}$, $\ln(|\cos b|) \to -\infty$.
So, the limit is $-(-\infty) = +\infty$.
Conclusion: The integral $\int_{0}^{\pi/2} \tan x \, dx$ diverges.


Comparing the results for the four options:
(A) Diverges
(B) Converges
(C) Diverges
(D) Diverges

The only convergent integral is option (B).
$$\boxed{{\displaystyle (B)\hspace{0.17em}{\int }_0^1\frac{1}{\sqrt[3]{x}}dx}}$$
\boxed{ (B) \, \int_{0}^{1} \frac{1}{\sqrt[3]{x}} dx }

A function $f(x)$ is continuous at $x = 0$ if:

$$\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }f(x)=f(0)$$
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)




$$f(0)=\cos (0-\pi k)=\cos (\pi k)$$
f(0) = \cos(0 - \pi k) = \cos(\pi k)




$$\underset{x\to 0^{-}}{\lim }f(x)=\cos (\pi k)$$
\lim_{x \to 0^-} f(x) = \cos(\pi k)




Using Taylor expansions:

$$\sin (x)=x-\frac{x^3}{6}+\dots ,\cos (x)=1-\frac{x^2}{2}+\dots$$
\sin(x) = x - \frac{x^3}{6} + \dots, \quad \cos(x) = 1 - \frac{x^2}{2} + \dots


$$\frac{\sin (x)-x}{1-\cos (x)}=\frac{-\frac{x^3}{6}}{\frac{x^2}{2}}=\frac{-x}{3}$$
\frac{\sin(x) - x}{1 - \cos(x)} = \frac{-\frac{x^3}{6}}{\frac{x^2}{2}} = \frac{-x}{3}


Thus:

$$\underset{x\to 0^{+}}{\lim }f(x)=0$$
\lim_{x \to 0^+} f(x) = 0




$$\cos (\pi k)=0$$
\cos(\pi k) = 0




$$\pi k=\frac{\pi }{2},\frac{3\pi }{2},\dots \Rightarrow k=\frac{1}{2},\frac{3}{2},\dots$$
\pi k = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \quad \Rightarrow \quad k = \frac{1}{2}, \frac{3}{2}, \dots


The smallest positive value is:

$$k=\frac{1}{2}$$
k = \frac{1}{2}




$$\boxed{{\displaystyle (B)\hspace{0.17em}k=\frac{1}{2}}}$$
\boxed{(B) \, k = \frac{1}{2}}