Math / Calculus

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

Studdy Solution

STEP 1

What is this asking? We need to compute a definite integral of a function involving $x$ raised to negative powers. Watch out! Don't forget about the power rule for integration, and be careful with those negative exponents!
Also, remember to apply the bounds correctly.

STEP 2

1. Rewrite the integrand
2. Integrate
3. Evaluate

STEP 3

Alright, let's rewrite this integral to make it easier to work with!
We can rewrite $\frac{1}{x^3}$ as $x^{-3}$ and $\frac{2}{x^2}$ as $2x^{-2}$ .
So, our integral becomes: $\int_{1}^{4} (x^{-3} - 2x^{-2}) \, dx$ This makes it much clearer for applying the power rule in the next step!

STEP 4

Now, let's integrate term by term using the power rule.
Remember, the power rule for integration says: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ , where $n \neq -1$ .
Don't worry, in our case, $n$ is $-3$ and $-2$ , so we're good!

STEP 5

Applying the power rule to $x^{-3}$ , we get $\frac{x^{-3+1}}{-3+1} = \frac{x^{-2}}{-2}$ .

STEP 6

Applying the power rule to $-2x^{-2}$ , we get $-2 \cdot \frac{x^{-2+1}}{-2+1} = -2 \cdot \frac{x^{-1}}{-1} = 2x^{-1}$ .

STEP 7

Putting it all together, our integral becomes: $\left[ \frac{x^{-2}}{-2} + 2x^{-1} \right]_1^4$ Remember, we don't need the constant of integration $C$ for definite integrals!

STEP 8

Time to evaluate!
Let's plug in our upper bound, which is 4, and our lower bound, which is 1.

STEP 9

Plugging in $x = 4$ , we get: $\frac{4^{-2}}{-2} + 2 \cdot 4^{-1} = \frac{1}{-2 \cdot 16} + \frac{2}{4} = -\frac{1}{32} + \frac{1}{2}$

STEP 10

Plugging in $x = 1$ , we get: $\frac{1^{-2}}{-2} + 2 \cdot 1^{-1} = -\frac{1}{2} + 2$

STEP 11

Now, let's subtract the result from the lower bound from the result of the upper bound: $\left( -\frac{1}{32} + \frac{1}{2} \right) - \left( -\frac{1}{2} + 2 \right) = -\frac{1}{32} + \frac{1}{2} + \frac{1}{2} - 2 = -\frac{1}{32} + 1 - 2 = -\frac{1}{32} - 1 = -\frac{1}{32} - \frac{32}{32} = -\frac{33}{32}$

STEP 12

Our final answer is $-\frac{33}{32}$ .

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