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Aloiza [94]
3 years ago
8

Evaluate the integral ∫2−1|x−1|dx

Mathematics
1 answer:
defon3 years ago
5 0

I think you might be referring to the definite integral,

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx

Recall the definition of absolute value:

|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

Then |x-1|=x-1 if x\ge1, and |x-1|=1-x is x. So spliting up the integral at <em>x</em> = 1, we have

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx

The rest is simple:

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}

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If the denominator does not equal 0, the equivalent expression of \frac{14x^4y^6}{7x^8y^2} is \frac{2y^{4}}{x^{4}}\frac{x^{10} y^{14}}{729}

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The expression is given as:

\frac{14x^4y^6}{7x^8y^2}

Divide 14 by 7

\frac{14x^4y^6}{7x^8y^2} = \frac{2x^4y^6}{x^8y^2}

Apply the law of indices

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Hence, the equivalent expression of \frac{14x^4y^6}{7x^8y^2} is \frac{2y^{4}}{x^{4}}

Read more about equivalent expressions at:

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brainlest if it helped

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