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Evaluate the following iterated integrals. $$\int_{1}^{2} \int_{0}^{1}\left(3 x^{2}+4 y^{3}\right) d y d x$$

Short Answer

Expert verified
Question: Evaluate the iterated integral $$\int_{1}^{2} \int_{0}^{1} (3x^2 + 4y^3) dy dx$$. Answer: 8

Step by step solution

01

Identify the inner integral bounds and the expression

First, we need to identify the expression with respect to y and its bounds. The expression is: $$3x^2 + 4y^3$$ and the bounds are from 0 to 1, i.e., y varies from 0 to 1.
02

Integrate with respect to y

Now, we will integrate the expression \((3x^2 + 4y^3)\) with respect to y from 0 to 1: $$\int_{0}^{1} (3x^2 + 4y^3) dy$$ This integral becomes: $$\int_{0}^{1} (4y^3) dy + \int_{0}^{1} (3x^2) dy$$
03

Evaluate the inner integral

Evaluating the above integrals, we get the following expressions: $$\left[\frac{4}{4}y^4\right]_0^1 + \left[3x^2 y\right]_0^1$$ After evaluating the limits, we get: $$1^4 - 0^4 + 3x^2(1) - 3x^2(0) = 1 + 3x^2$$
04

Integrate with respect to x

Now, we will integrate the expression \((\)1 + 3x^2$$)$ with respect to x from 1 to 2: $$\int_{1}^{2} (1 + 3x^2) dx$$
05

Evaluate the outer integral

Evaluating the above integral, we get the following expression: $$\left[x + \frac{3}{3} x^3\right]_1^2$$ After evaluating the limits, we get: $$(2 + 2^3) - (1 + 1^3) = 2 + 8 -1 -1 = 8$$ Thus, the value of the given iterated integral is 8.

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