Calculus 1 Review

Solution for Question 1

Here is the step-by-step breakdown for solving each part of the calculus problem based on the function:

(a) Find $f'(x)$

Concept: The first derivative represents the instantaneous rate of change or the slope of the function at any given point. To find it, we use the Power Rule ($\frac{d}{dx}[x^n] = n x^{n-1}$) and the Constant Rule ($\frac{d}{dx}[c] = 0$).

1. Differentiate each term of the function individually:

• The derivative of $x^2$ is $2x$.

• The derivative of $-4x$ is $-4$.

• The derivative of the constant $4$ is $0$.

2. Combine the terms to get your derivative function:

(b) Find an equation of the line tangent to the graph of $f(x)$ at $x = -1$

Concept: To write the equation of a straight line, you need a point $(x_1, y_1)$ and a slope ($m$). Once you have both, you plug them into the point-slope formula: $y - y_1 = m(x - x_1)$.

1. Find the y-coordinate (the point): Evaluate the original function $f(x)$ at $x = -1$.

This gives you the coordinate point $(-1, 9)$.

2. Find the slope ($m$): Evaluate the derivative function $f'(x)$ at $x = -1$.

3. Construct the equation: Substitute the point $(-1, 9)$ and slope $m = -6$ into the point-slope form.

(c) For what value(s) of $x$ does $f(x)$ have a slope of 0?

Concept: Because the derivative $f'(x)$ gives you the slope at any $x$, you can find where the slope is exactly $0$ by setting the derivative equal to zero and solving for $x$.

1. Set $f'(x) = 0$:

2. Isolate $x$:

(d) For what value(s) of $x$ does $f(x)$ have a negative instantaneous rate of change?

Concept: "Instantaneous rate of change" is simply another term for the derivative. For this rate to be negative, the derivative must be less than zero ($f'(x) < 0$).

1. Set up the inequality using your derivative:

2. Solve for $x$:

In interval notation, the instantaneous rate of change is negative on the interval $(-\infty, 2)$.

Before we dive in, remember the golden rule of indefinite integrals: never forget to add the constant of integration ($+ C$) at the very end!

(a) $\int \left(x + \frac{2}{x} - \frac{3}{x^3} + \frac{4}{\sqrt[4]{x}}\right) dx$

Step 1: Rewrite the terms using exponents Just like with derivatives, it is much easier to integrate fractions and roots if we convert them into standard power forms ($x^n$).

• $\frac{2}{x}$ stays as $2 \cdot \frac{1}{x}$ because it follows a special rule.

• $-\frac{3}{x^3}$ becomes $-3x^{-3}$

• $\frac{4}{\sqrt[4]{x}}$ becomes $4x^{-1/4}$

Now, rewrite the whole integral:

Step 2: Integrate term-by-term We will use the Power Rule for Integration ($\int x^n dx = \frac{x^{n+1}}{n+1}$) for most of these, and the Natural Log Rule ($\int \frac{1}{x} dx = \ln|x|$) for the second term.

• First term ($x^1$): Add 1 to the exponent, then divide by the new exponent.

• Second term ($2 \cdot \frac{1}{x}$): The integral of $\frac{1}{x}$ is always $\ln|x|$.

• Third term ($-3x^{-3}$): Add 1 to the exponent ($-3 + 1 = -2$), then divide by $-2$.

• Fourth term ($4x^{-1/4}$): Add 1 to the exponent ($-\frac{1}{4} + 1 = \frac{3}{4}$), then divide by $\frac{3}{4}$ (which is the same as multiplying by $\frac{4}{3}$).

(b) $\int (2x + 1)\sqrt[5]{x^2 + x} \, dx$

Step 1: Identify the method ($u$-substitution) Notice that we have a complicated "inner" function ($x^2 + x$) sitting inside a fifth root, and its exact derivative ($2x + 1$) is multiplying the rest of the expression. This is a textbook cue to use $u$-substitution.

Let's define our pieces:

• Let $u = x^2 + x$

• Take the derivative to find $du$: $du = (2x + 1) \, dx$

Step 2: Substitute $u$ and $du$ into the integral Replace $(x^2 + x)$ with $u$, and replace $(2x + 1)dx$ with $du$.

Step 3: Rewrite and integrate Convert the root into a fractional exponent so you can apply the Power Rule:

Add 1 to the exponent ($\frac{1}{5} + 1 = \frac{6}{5}$), and divide by $\frac{6}{5}$ (multiply by $\frac{5}{6}$):

Step 4: Substitute back to the original variable ($x$) Replace $u$ back with your original expression $(x^2 + x)$ to finish the job.

Final Answer:

Unlike indefinite integrals, definite integrals give us a specific numerical value (representing the net area under the curve between two boundaries) and do not need a $+ C$ at the end. We will use the Fundamental Theorem of Calculus:

where $F(x)$ is the antiderivative.

(a) $\int_{1}^{3} \left(3x^2 - \frac{1}{4}x^3\right) dx$

Step 1: Find the antiderivative $F(x)$ Integrate each term individually using the Power Rule ($\int x^n dx = \frac{x^{n+1}}{n+1}$):

• First term ($3x^2$): $\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3$

• Second term ($-\frac{1}{4}x^3$): $\int -\frac{1}{4}x^3 \, dx = -\frac{1}{4} \cdot \frac{x^4}{4} = -\frac{1}{16}x^4$

So, our antiderivative function is:

Step 2: Evaluate at the upper limit ($x = 3$) Plug $3$ into your antiderivative:

Step 3: Evaluate at the lower limit ($x = 1$) Plug $1$ into your antiderivative:

Step 4: Subtract the lower bound value from the upper bound value ($F(3) - F(1)$)

Simplifying the fraction gives exactly 21.

(b) $\int_{0}^{2} \frac{1}{5x+1} dx$

Step 1: Use $u$-substitution to find the antiderivative Because the denominator contains a linear function ($5x + 1$), we need to use a simple $u$-sub.

• Let $u = 5x + 1$

• Take the derivative: $du = 5 \, dx \implies dx = \frac{1}{5} \, du$

Step 2: Change the limits of integration to match $u$ When dealing with definite integrals, it's easiest to convert the $x$-boundaries into $u$-boundaries using your substitution formula ($u = 5x + 1$):

• Lower limit ($x = 0$): $u = 5(0) + 1 = 1$

• Upper limit ($x = 2$): $u = 5(2) + 1 = 11$

Step 3: Rewrite and evaluate the integral in terms of $u$ Substitute your components and the new boundaries into the integral:

The integral of $\frac{1}{u}$ is $\ln|u|$:

Step 4: Apply the boundaries

Since $\ln(1) = 0$, this simplifies beautifully:

Final Answer: