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Step-by-step Solution

Integrate $y+x^2-6-\left(y+2x-3\right)$ from $-3$ to $5$

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Answer

$10.6667$

Step-by-step explanation

Problem to solve:

$\int_{\left(-3\right)}^5\left[\left(y+x^2-6\right)-\left(y+2x-3\right)\right]dx$
1

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int_{-3}^{5} ydx+\int_{-3}^{5} x^2dx+\int_{-3}^{5}-6dx+\int_{-3}^{5}\left(-y-2x+3\right)dx$
2

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int_{-3}^{5} ydx+\int_{-3}^{5} x^2dx+\int_{-3}^{5}-6dx+\int_{-3}^{5}-ydx+\int_{-3}^{5}-2xdx+\int_{-3}^{5}3dx$

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Answer

$10.6667$
$\int_{\left(-3\right)}^5\left[\left(y+x^2-6\right)-\left(y+2x-3\right)\right]dx$

Main topic:

Definite integrals

Used formulas:

3. See formulas

Time to solve it:

~ 0.94 seconds