Practice Exam 4 – MTH 141

Problem 1: Evaluate the following integral:

Answer: 104 1/6

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!

Answer: 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:

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