# Step-by-step Solution

## Integrate y+x^2-6-(y+2x-3) from -3 to 5

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$\frac{32}{3}$

## 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

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

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

$\int_{-3}^{5} ydx+\int_{-3}^{5} x^2dx+\int_{-3}^{5}-6dx+\int_{-3}^{5}-ydx+\int_{-3}^{5}-2xdx+\int_{-3}^{5}-1\left(-3\right)dx$

$\frac{32}{3}$

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

### Main topic:

Integral calculus

~ 0.95 seconds