Free Single Variable Calculus Course Materials Study Resources

To find the derivative of an integral with a variable upper limit, we apply the First Fundamental Theorem of Calculus combined with the Chain Rule.

The general formula is: $\frac{d}{dx} \int_a^{g(x)} f(t) \, dt = f(g(x)) \cdot g'(x)$

Step 1: Identify the components

• Integrand $f(t) = \ln(t^2 + 1)$
• Upper limit $g(x) = x^2$
• Lower limit $a = 2$ (Since this is a constant, its derivative is 0 and does not affect the result)

Step 2: Apply the components to the formula First, evaluate the integrand $f(t)$ at the upper limit $g(x)$: $f(g(x)) = \ln((x^2)^2 + 1) = \ln(x^4 + 1)$

Next, find the derivative of the upper limit $g'(x)$: $g'(x) = \frac{d}{dx}(x^2) = 2x$

Step 3: Multiply to find the final derivative $f(g(x)) \cdot g'(x) = \ln(x^4 + 1) \cdot 2x$

Rearranging the terms, we get: $\displaystyle 2x \ln(x^4 + 1)$

To find the radius and interval of convergence, we follow these steps:

1. Apply the Ratio Test: We calculate the limit $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$. $L = \lim_{n \to \infty} \left| \frac{(n+1)(x+2)^{n+1}}{4^{n+1} ((n+1)^2+1)} \cdot \frac{4^n (n^2+1)}{n(x+2)^n} \right|$ Simplifying the expression: $L = \frac{|x+2|}{4} \lim_{n \to \infty} \left| \frac{(n+1)(n^2+1)}{n(n^2+2n+2)} \right| = \frac{|x+2|}{4} \cdot 1 = \frac{|x+2|}{4}$

2. Determine the Radius ($R$): The series converges when $L < 1$: $\frac{|x+2|}{4} < 1 \implies |x+2| < 4$ Thus, the radius of convergence $R = 4$. This gives the open interval $-4 < x+2 < 4$, which is $-6 < x < 2$.

3. Test Endpoint $x = 2$: Substitute $x=2$ into the original series: $\sum_{n=1}^{\infty} \frac{n(2+2)^n}{4^n (n^2+1)} = \sum_{n=1}^{\infty} \frac{n \cdot 4^n}{4^n (n^2+1)} = \sum_{n=1}^{\infty} \frac{n}{n^2+1}$ Using the Limit Comparison Test with the harmonic series $\sum \frac{1}{n}$: $\lim_{n \to \infty} \frac{n/(n^2+1)}{1/n} = \lim_{n \to \infty} \frac{n^2}{n^2+1} = 1$ Since $\sum \frac{1}{n}$ diverges, the series diverges at $x = 2$.

4. Test Endpoint $x = -6$: Substitute $x=-6$ into the original series: $\sum_{n=1}^{\infty} \frac{n(-6+2)^n}{4^n (n^2+1)} = \sum_{n=1}^{\infty} \frac{n(-4)^n}{4^n (n^2+1)} = \sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2+1}$ This is an alternating series. By the Alternating Series Test (AST), where $b_n = \frac{n}{n^2+1}$:

• $b_n > 0$ for all $n \ge 1$.
• $\lim_{n \to \infty} \frac{n}{n^2+1} = 0$.
• $b_n$ is decreasing for $n \ge 1$ (checked via derivative $f'(x) = \frac{1-x^2}{(x^2+1)^2} < 0$). The series converges at $x = -6$.

Combining these results, the interval of convergence is $[-6, 2)$.

To determine the maximum error bound, we use Taylor's Theorem with the Lagrange form of the remainder for $n=2$.

1. Identify the Remainder Formula: For a second-degree polynomial ($n=2$), the remainder $R_2(x)$ is given by: $R_2(x) = \frac{f^{(3)}(c)}{3!}(x-a)^3$ where $c$ is some value between the center $a$ and the point of interest $x$.

2. Calculate the Third Derivative: Find the successive derivatives of $f(x) = \cos(x)$:

• $f'(x) = -\sin(x)$
• $f''(x) = -\cos(x)$
• $f^{(3)}(x) = \sin(x)$

1. Determine the Maximum of the Derivative: We are approximating at $x = \pi/6$ with center $a = 0$, so $c \in [0, \pi/6]$. The function $|f^{(3)}(c)| = |\sin(c)|$ is monotonically increasing on this interval. Therefore, its maximum value occurs at the upper bound $c = \pi/6$: $|f^{(3)}(c)| \leq \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

2. Calculate the Error Bound: Substitute the maximum derivative value, $x = \pi/6$, and $3! = 6$ into the remainder formula: $|R_2(\pi/6)| \leq \frac{1/2}{6} \left(\frac{\pi}{6}\right)^3$ $|R_2(\pi/6)| \leq \frac{1}{12} \cdot \frac{\pi^3}{216} = \frac{\pi^3}{2592}$

The final answer is A.

To find the derivative of an accumulation function where the upper limit is a function of $x$, we apply the Fundamental Theorem of Calculus (FTC) Part 1 combined with the Chain Rule.

1. Define the Rule: The rule states that if $F(x) = \int_{a}^{g(x)} f(t) \, dt$, then $F'(x) = f(g(x)) \cdot g'(x)$.
2. Identify the Components:
   • The integrand is $f(t) = \sqrt{1 + t^2}$.
   • The upper limit is $g(x) = x^3$.
   • The lower limit is a constant ($a = 2$).
3. Apply the Rule:
   • First, substitute the upper limit into the integrand: $f(g(x)) = \sqrt{1 + (x^3)^2} = \sqrt{1 + x^6}$.
   • Next, multiply by the derivative of the upper limit: $g'(x) = \frac{d}{dx}(x^3) = 3x^2$.
4. Combine: $F'(x) = \sqrt{1 + x^6} \cdot 3x^2 = 3x^2\sqrt{1 + x^6}$.

Therefore, the final answer is $3x^2\sqrt{1 + x^6}$.

To determine the possible range of values for $f(5)$, we apply the Mean Value Theorem (MVT). The MVT states that for a function $f$ continuous on $[a, b]$ and differentiable on $(a, b)$, there exists at least one point $c \in (a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

In this problem, $a = 1$, $b = 5$, and $f(1) = 7$. Rearranging the MVT formula to solve for $f(5)$ gives:

$f(5) = f(1) + f'(c)(5 - 1)$ $f(5) = 7 + 4f'(c)$

We are given the bounds for the derivative: $-2 \le f'(x) \le 3$. Since $c$ is a point in the interval $(1, 5)$, $f'(c)$ must also satisfy these bounds:

1. Lower Bound: To find the minimum possible value for $f(5)$, we use the minimum value of $f'(c)$: $f(5) \ge 7 + 4(-2) = 7 - 8 = -1$

2. Upper Bound: To find the maximum possible value for $f(5)$, we use the maximum value of $f'(c)$: $f(5) \le 7 + 4(3) = 7 + 12 = 19$

Combining these results, we find that the range of possible values for $f(5)$ is $-1 \le f(5) \le 19$.

To find the absolute extrema of a continuous function on a closed interval $[a, b]$, follow the Closed Interval Method:

1. Find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 1) = 3x^2 - 6x$

2. Identify critical points by setting $f'(x) = 0$: $3x(x - 2) = 0$ The solutions are $x = 0$ and $x = 2$.

3. Filter critical points for the interval $[-1, 1]$:

   • $x = 0$ is within the interval $[-1, 1]$.
   • $x = 2$ is not within the interval and is excluded from consideration.

4. Evaluate the function at the critical point and the endpoints:

   • At $x = 0$: $f(0) = (0)^3 - 3(0)^2 + 1 = 1$
   • At left endpoint $x = -1$: $f(-1) = (-1)^3 - 3(-1)^2 + 1 = -3$
   • At right endpoint $x = 1$: $f(1) = (1)^3 - 3(1)^2 + 1 = -1$

5. Identify extrema: Comparing the values $\{1, -3, -1\}$, the maximum value is 1 and the minimum value is $-3$.

The absolute maximum is 1 and the absolute minimum is -3.

The derivative is defined as the instantaneous rate of change of a function.

1. Start with the Difference Quotient: The expression $\frac{f(x+h) - f(x)}{h}$ represents the slope of the secant line between the points $(x, f(x))$ and $(x+h, f(x+h))$. This is the average rate of change over the interval $h$.
2. Apply the Limit: To find the slope of the tangent line (the derivative) at exactly $x$, we must let the distance $h$ between the two points approach zero. This corresponds to the limit $\lim_{h \to 0}$.
3. Combine: Putting these together, the formal definition is: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Analysis of Distractors:

• Option A is the difference quotient (average rate of change) without the limit.
• Option C takes the limit as $h \to \infty$, which is incorrect for finding a local slope.
• Option D has an incorrect sum ($+$) in the numerator instead of a difference ($-$).

Therefore, the correct expression is Option B.

Step 1: Determine the intersection points To find the limits of integration, set the functions equal to each other: $5x - x^2 = x$ $4x - x^2 = 0$ $x(4 - x) = 0$ The curves intersect at $x = 0$ and $x = 4$. These are our limits of integration $a=0$ and $b=4$.

Step 2: Identify the upper and lower curves Pick a test point in the interval $(0, 4)$, such as $x = 1$:

• $y_{parabola} = 5(1) - (1)^2 = 4$
• $y_{line} = 1$

Since $4 > 1$, the parabola$y = 5x - x^2 is the upper function (f(x)$) and the line$y = x is the lower function (g(x)$).

Step 3: Set up the definite integral The area $A$ is given by the integral of the upper curve minus the lower curve: $A = \int_{0}^{4} [(5x - x^2) - x] \, dx$ $A = \int_{0}^{4} (4x - x^2) \, dx$

Step 4: Evaluate the integral Apply the Power Rule for integration: $A = \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4}$

Substitute the upper limit ($x=4$) and lower limit ($x=0$): $A = \left( 2(4)^2 - \frac{4^3}{3} \right) - \left( 2(0)^2 - \frac{0^3}{3} \right)$ $A = \left( 32 - \frac{64}{3} \right) - 0$ $A = \frac{96}{3} - \frac{64}{3}$ $A = \mathbf{\frac{32}{3}}$

To find the derivative of an accumulation function where the upper limit is a function of $x$, we use the Fundamental Theorem of Calculus (Part 1) combined with the Chain Rule.

The theorem states: $\frac{d}{dx} \left[ \int_{a}^{u(x)} f(t) \, dt \right] = f(u(x)) \cdot u'(x)$

Step 1: Identify the components

• The integrand function is $f(t) = \frac{1}{t^2 + 1}$.
• The upper limit is $u(x) = x^2$.
• The lower limit $a = 1$ is a constant (so its derivative is 0).

Step 2: Compose the function $f(u(x))$ Substitute $u(x) = x^2$ into $f(t)$: $f(u(x)) = f(x^2) = \frac{1}{(x^2)^2 + 1} = \frac{1}{x^4 + 1}$

Step 3: Find the derivative of the upper limit $u'(x)$ $u'(x) = \frac{d}{dx}(x^2) = 2x$

Step 4: Apply the formula Multiply the composed function by the derivative of the upper limit: $g'(x) = f(u(x)) \cdot u'(x)$ $g'(x) = \left( \frac{1}{x^4 + 1} \right) \cdot (2x)$ $g'(x) = \frac{2x}{x^4 + 1}$

Therefore, the correct derivative is $\frac{2x}{x^4 + 1}$.

To classify the discontinuities, we must evaluate the limits and function values at the points of interest. Step 1: Analyze $f(x)$ at $x=1$. Substituting $x=1$ into $f(x) = \frac{x^2 - 4x + 3}{x^2 - 1}$ results in the indeterminate form $\frac{0}{0}$. Factoring the numerator and denominator gives $f(x) = \frac{(x-1)(x-3)}{(x-1)(x+1)}$. For $x \neq 1$, the function simplifies to $f(x) = \frac{x-3}{x+1}$. The limit is $\lim_{x \to 1} f(x) = \frac{1-3}{1+1} = -1$. Because the limit exists but the function is undefined at $x=1$, $f(x)$ has a removable discontinuity. Step 2: Analyze $g(x)$ at $x=0$. We evaluate the one-sided limits. The left-hand limit is $\lim_{x \to 0^-} \frac{\tan(2x)}{x} = \lim_{x \to 0^-} 2 \cdot \frac{\tan(2x)}{2x} = 2(1) = 2$. The right-hand limit is $\lim_{x \to 0^+} 1 = 1$. Because both one-sided limits are finite but not equal ($2 \neq 1$$), g(x)$ has a jump discontinuity. Comparing the results, the correct pair is removable and jump.

To evaluate the integral $\int x \cos(x^2) \, dx$, we use the Substitution Rule.

Step 1: Identify the substitution. Let $u = x^2$ be the inside function.

Step 2: Find the differential $du$. Differentiating $u$ with respect to $x$: $du = 2x \, dx$

Step 3: Adjust for the constant. The integrand contains $x \, dx$, not $2x \, dx$. We can solve for $x \, dx$ by dividing both sides by 2: $x \, dx = \frac{1}{2} \, du$

Step 4: Substitute into the integral. Replace $x^2$ with $u$ and $x \, dx$ with $\frac{1}{2} \, du$: $\int \cos(u) \cdot \frac{1}{2} \, du = \frac{1}{2} \int \cos(u) \, du$

Step 5: Integrate. The antiderivative of $\cos(u)$ is $\sin(u)$: $\frac{1}{2} \sin(u) + C$

Step 6: Substitute back. Replace $u$ with the original function $x^2$: \textbf{\frac{1}{2} \sin(x^2) + C}

To determine differentiability at $x=0$, we must evaluate the limit of the difference quotient, as standard rules like the product rule require differentiability of individual factors (which $|x|$ lacks at $x=0$).

The derivative is defined as: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$

Step 1: Substitute the function. We have $f(0) = 0|0| = 0$ and $f(h) = h|h|$. Substituting these into the limit: $f'(0) = \lim_{h \to 0} \frac{h|h| - 0}{h}$

Step 2: Simplify the expression. For $h \neq 0$, we can cancel $h$ in the numerator and denominator: $\frac{h|h|}{h} = |h|$

Step 3: Evaluate the limit. Now we evaluate the limit of the simplified expression: $f'(0) = \lim_{h \to 0} |h|$

As $h$ approaches 0 from either the positive or negative side, $|h|$ approaches 0.

Since the limit exists and equals 0, $f(x)$ is differentiable at $x=0$ and $f'(0) = 0$.

To find the derivative of the rational function $h(x) = \frac{x^2 + 1}{x - 2}$, we apply the Quotient Rule:

$h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

where $f(x) = x^2 + 1$ and $g(x) = x - 2$.

1. Find the derivatives of the numerator and denominator:

   • $f'(x) = \frac{d}{dx}(x^2 + 1) = 2x$
   • $g'(x) = \frac{d}{dx}(x - 2) = 1$

2. Substitute these into the Quotient Rule formula: $h'(x) = \frac{(2x)(x - 2) - (x^2 + 1)(1)}{(x - 2)^2}$

3. Simplify the numerator by distributing and combining like terms:

   • Numerator: $2x^2 - 4x - (x^2 + 1) = 2x^2 - 4x - x^2 - 1$
   • Simplified Numerator: $x^2 - 4x - 1$

Therefore, the derivative is $\frac{x^2 - 4x - 1}{(x - 2)^2}$.

The definite integral $\int_{a}^{b} f(x) \, dx$ is formally defined as the limit of a Riemann sum: $\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$ Geometrically, this sum represents the net signed area between the curve and the $x$-axis. Regions where the function is above the $x$-axis ($f(x) > 0$) contribute positive area, while regions below the $x$-axis ($f(x) < 0$) contribute negative area. Option A describes the derivative, Option C describes the indefinite integral, and Option D describes the integral of $|f(x)|$. The correct definition is B.

To calculate the derivative of $h(x) = (3x^2 + 5)^4$, we use the Chain Rule, which states that if $h(x) = f(g(x))$, then $h'(x) = f'(g(x)) \cdot g'(x)$.

1. Identify the inner and outer functions:

   • Outer function: $f(u) = u^4$
   • Inner function: $g(x) = 3x^2 + 5$

2. Differentiate each function:

   • The derivative of the outer function with respect to $u$ is $f'(u) = 4u^3$.
   • The derivative of the inner function with respect to $x$ is $g'(x) = 6x$.

3. Apply the Chain Rule: $h'(x) = f'(g(x)) \cdot g'(x)$ $h'(x) = 4(3x^2 + 5)^3 \cdot (6x)$

4. Simplify the expression: $h'(x) = 24x(3x^2 + 5)^3$

The final derivative is $24x(3x^2 + 5)^3$.