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Step-by-step Solution

Integrate $2x-x^2-x^2$ from $0.25$ to $0.75$

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Answer

$0.2292$

Step-by-step explanation

Problem to solve:

$\int_{\frac{1}{4}}^{\frac{3}{4}}\left(2x-x^2-x^2\right)dx$
1

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

$\int_{\frac{1}{4}}^{\frac{3}{4}}2xdx+\int_{\frac{1}{4}}^{\frac{3}{4}}-x^2dx+\int_{\frac{1}{4}}^{\frac{3}{4}}-x^2dx$
2

Adding $\int_{\frac{1}{4}}^{\frac{3}{4}}-x^2dx$ and $\int_{\frac{1}{4}}^{\frac{3}{4}}-x^2dx$

$\int_{\frac{1}{4}}^{\frac{3}{4}}2xdx+2\int_{\frac{1}{4}}^{\frac{3}{4}}-x^2dx$

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Answer

$0.2292$
$\int_{\frac{1}{4}}^{\frac{3}{4}}\left(2x-x^2-x^2\right)dx$

Main topic:

Definite integrals

Used formulas:

2. See formulas

Time to solve it:

~ 0.78 seconds