 # MTH 141 Practice Problem Bank

Problem 1: Evaluate the following limits. When appropriate, declare whether the limit goes to ∞ or -∞ if the limit does not exist. PAID CONTENT This is the end of the preview. To unlock the rest, get the .

Problem 1: Evaluate the derivative for the following functions. PAID CONTENT This is the end of the preview. To unlock the rest, get the .

Problem 1: Evaluate the following integrals:

To evaluate this indefinite integral, we'll rely on the reverse power rule and the sum-difference rule for integration.

Let’s start by applying the sum-difference rule to simplify the integral more.

Now that we've separated each term into its own integral, we can evaluate all of these integrals by applying the reverse power rule.

Remember for the reverse-power rule, you first increase the exponent of x by 1 and then divide by that number. To apply the reverse-power rule to the first term here...

...we'll complete the following process:

For the second term...

...we'll do the following:

And for the last term...

...we've gotta be careful: if you apply the normal reverse-power rule principles to the example, then you’re left with undefined, which doesn’t work for us. We must remember on of our identities for this integral!

The result is always the natural log of the absolute value of x. As you may see in future problems, the term in the absolute value may change, but you will always need the natural log and absolute value signs, as the absolute value signs are used to prevent any negative numbers from being input into the natural log expression.

When we combine all these reverse-power rule outputs together, we get the following as our final answer!

Remember, the +C term is crucial for indefinite integral expressions. This is to generalize our solution, as any constant could be substituted in for C and still give us a valid answer.

Before you jump the gun and start using complex integration techniques, I encourage you to look at this problem more closely.

One of the best things you can do before you integrate any expression is see if the expression can be simplified algebraically or trigonometrically.

In this case, we can simplify this expression to get rid of the fraction!  Let’s do this by expanding the fraction as such…

...and then further cancel out x terms in the denominator like so!

From here, this is a piece-of-cake. We've just gotta apply the reverse-power rule to each term! Let's start with the first term...

...which would look like so:

The second term...

...would yield the following output with the reverse-power rule:

When we combine all these reverse-power rule outputs together, we get the following as our final answer! PAID CONTENT This is the end of the preview. To unlock the rest, get the .

Problem 1: Evaluate the following integral:

Remember: when faced with a definite integral (one with bounds)...

...we'll start by evaluating as if it was an indefinite integral, and then plug in the bound values to determine our answer!

### Evaluate as an indefinite integral

To evaluate this as if it were an indefinite integral, we can start by identifying the sum rule applies here. In other words, we can take the integral of each of these terms individually, starting with x2!

We'll apply the reverse power rule to x2 like so...

...and then simplify to the following:

Next, we apply the reverse power rule to 3x like so...

...and simplify to the following:

And lastly, we can apply the reverse power rule to 5 (remember, there's an x0 next to 5!) like so...

...and simplify to the following:

Now, it's time to plug in the bounds!

### Plug-in bound values

When evaluating definite integral, we must lean on the following principle:

We solved for F(x) here...

...and have defined our upper-bound (of 5) and lower-bound (of 0) here...

...meaning that to solve this definite integral, we need to do the following:

Let's solve for F(5) first. When we plug in x = 5 to F(x)...

...we get the following value:

Next, let's solve for F(0). When we plug in x = 0 to F(x)...

...we get the following value:

When we plug F(5) = 104 1/6 and F(0) = 0 into the following...

...we get a final answer of 104 1/6!

Problem 2: Evaluate the following partial derivative:

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Remember for partial derivatives, we want to only focus on taking the derivative of the variable specifically identified in the derivative operator symbol.

In this case, that's x!

So, let's take the derivative in x-terms and neglect the other variables (y & z).

We'll start by applying the difference rule to separate these terms.

Now, let's take the derivative of each, applying the power rule. We'll start with x2y.

Remember: we're only taking derivative with respect to x here. Any other variables (y, in this case), should be treated like a constant.

Therefore, we can move y out to the front (like we would a normal constant)...

...and apply the power rule to x2 like so!

Which, when simplified, results in the following: PAID CONTENT This is the end of the preview. To unlock the rest, get the . PAID CONTENT 