Math / Calculus

QuestionProblem 1. (1 point) $\int_{0}^{2} \int_{0}^{5} e^{5 x+2 y} d x d y=\square$ (Give the exact value, not a decimal approximation)

Studdy Solution

STEP 1

What is this asking? We need to calculate the exact value of a double integral of $e^{5x+2y}$ over a rectangular region. Watch out! Don't forget to integrate with respect to both $x$ and $y$ , and apply the limits of integration correctly for each variable.

STEP 2

1. Integrate with respect to $x$ .
2. Integrate with respect to $y$ .

STEP 3

Alright, let's start with the inner integral with respect to $x$ .
We're looking at $\int_{0}^{2} \int_{0}^{5} e^{5x+2y} dx dy$ .
Notice how we're treating $y$ as a constant in this step!

STEP 4

We have $\int_{0}^{5} e^{5x+2y} dx$ .
Remember, $e^{5x+2y}$ can be rewritten as $e^{5x} \cdot e^{2y}$ .
Since $e^{2y}$ is a constant with respect to $x$ , we can pull it out of the inner integral: $e^{2y} \int_{0}^{5} e^{5x} dx$ .
Now, the integral of $e^{5x}$ is $\frac{1}{5}e^{5x}$ .
So, we have $e^{2y} \cdot \left[ \frac{1}{5}e^{5x} \right]_0^5$ .

STEP 5

Let's plug in our limits of integration for $x$ : $e^{2y} \cdot \left( \frac{1}{5}e^{5 \cdot 5} - \frac{1}{5}e^{5 \cdot 0} \right) = e^{2y} \cdot \left( \frac{1}{5}e^{25} - \frac{1}{5}e^0 \right) = \frac{1}{5}e^{2y} \left(e^{25} - 1 \right)$ .
See how we simplified?

STEP 6

Now, we'll integrate our result from the previous steps with respect to $y$ .
Our integral looks like this: $\int_{0}^{2} \frac{1}{5}e^{2y} \left(e^{25} - 1 \right) dy$ .

STEP 7

We can pull the constants out: $\frac{1}{5}(e^{25} - 1) \int_{0}^{2} e^{2y} dy$ .
The integral of $e^{2y}$ is $\frac{1}{2}e^{2y}$ , so we have $\frac{1}{5}(e^{25} - 1) \cdot \left[ \frac{1}{2}e^{2y} \right]_0^2$ .

STEP 8

Let's plug in the limits for $y$ : $\frac{1}{5}(e^{25} - 1) \cdot \left( \frac{1}{2}e^{2 \cdot 2} - \frac{1}{2}e^{2 \cdot 0} \right) = \frac{1}{5}(e^{25} - 1) \cdot \left( \frac{1}{2}e^4 - \frac{1}{2} \right) = \frac{1}{10}(e^{25} - 1)(e^4 - 1)$ .
Boom!

STEP 9

The exact value of the double integral is $\frac{1}{10}(e^{25} - 1)(e^4 - 1)$ .

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