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int_(0)^(3)int_(0)^(-x+5)y^(2)-xdydx

$\int_{0}^{3} \int_{0}^{-x+5} y^{2}-x d y d x$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\int_{0}^{3} \int_{0}^{-x+5} y^{2}-x d y d x

Set up inner integral: First, we set up the inner integral with respect to $y$ , keeping $x$ as a constant. $\int_{0}^{-x+5} (y^2 - x) \, dy$
Integrate with respect to $y$ : Next, we integrate $y^2 - x$ with respect to $y$ . The integral of $y^2$ is $(\frac{1}{3})y^3$ , and the integral of $x$ with respect to $y$ is $xy$ (since $x$ is constant with respect to $y$ ). $\newline$ = $y^2 - x$ $0$ evaluated from $y^2 - x$ $1$ to $y^2 - x$ $2$
Plug in limits: Plugging in the limits of integration: $\newline$ $= \left[\left(\frac{1}{3}\right)(-x+5)^3 - x(-x+5)\right] - [0 - 0]$ $\newline$ $= \left(\frac{1}{3}\right)(-x+5)^3 + x^2 - 5x$
Set up outer integral: Now, set up the outer integral with respect to $x$ : $\int_{0}^{3} \left[\frac{1}{3}(-x+5)^3 + x^2 - 5x\right] dx$
Integrate each term: We integrate each term separately: $\newline$ For $(1/3)(-x+5)^3$ , expand and integrate term by term: $\newline$ $= \int_{0}^{3} (1/3)(-x+5)^3 \,dx$ $\newline$ $= (1/3)\int_{0}^{3} (-x+5)^3 \,dx$ $\newline$ $= (1/3)\left[(-1/4)(-x+5)^4\right]$ evaluated from $0$ to $3$ $\newline$ $= (1/3)\left[(-1/4)(2)^4 - (-1/4)(5)^4\right]$ $\newline$ $= (1/3)\left[(-1/4)(16) - (-1/4)(625)\right]$ $\newline$ $= (1/3)\left[-4 + 156.25\right]$ $\newline$ $= 50.75$

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