Math / Calculus

Question $\int_{0}^{1} \frac{4 r d r}{\sqrt{4+2 r^{2}}}$

Studdy Solution

STEP 1

What is this asking? We need to calculate the definite integral of a function with respect to $r$ from 0 to 1.
It looks a little scary, but we can totally handle it! Watch out! Don't forget to apply the limits of integration after finding the antiderivative.
Also, be careful with the square root – it can be tricky!

STEP 2

1. Simplify the integrand
2. U-Substitution
3. Integrate
4. Evaluate

STEP 3

Alright, let's rewrite our integral to make it easier to work with.
We can pull that 4 out front: $\int_{0}^{1} \frac{4r}{\sqrt{4+2r^2}} dr = 4 \int_{0}^{1} \frac{r}{\sqrt{4+2r^2}} dr$ See? Much cleaner already!

STEP 4

Now, let's use a clever trick called u-substitution.
Let's set $u = 4 + 2r^2$ .
This means $du = 4r \, dr$ .
We almost have $4r \, dr$ in our integral, but we're missing a factor of 2.

STEP 5

No worries!
We can multiply and divide by 2 to get what we need: $4 \int_{0}^{1} \frac{r}{\sqrt{4+2r^2}} dr = 4 \cdot \frac{1}{2} \int_{0}^{1} \frac{2 \cdot r}{\sqrt{4+2r^2}} dr = 2 \int_{0}^{1} \frac{4r}{\sqrt{4+2r^2}} dr$ Now we have $4r \, dr$ which is equal to $du$ !

STEP 6

Don't forget to change the limits of integration.
When $r = 0$ , $u = 4 + 2(0)^2 = 4$ .
When $r = 1$ , $u = 4 + 2(1)^2 = 6$ .
So our integral becomes: $2 \int_{4}^{6} \frac{1}{\sqrt{u}} du$ Much simpler, right?

STEP 7

Let's rewrite the integral one more time to make it super easy to integrate: $2 \int_{4}^{6} u^{-1/2} du$ Now we can use the power rule for integration!

STEP 8

Remember, the power rule says $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ .
In our case, $n = -\frac{1}{2}$ , so $n+1 = \frac{1}{2}$ .
Applying the power rule, we get: $2 \left[ \frac{u^{1/2}}{1/2} \right]_4^6 = 2 \left[ 2u^{1/2} \right]_4^6 = 4 \left[ \sqrt{u} \right]_4^6$

STEP 9

Time to plug in our limits of integration and find the final answer: $4 (\sqrt{6} - \sqrt{4}) = 4 (\sqrt{6} - 2) = 4\sqrt{6} - 8$ And there we have it!

STEP 10

The value of the definite integral is $4\sqrt{6} - 8$ .

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