Use Property 8 to estimate the value of the integral. ∫−111+x4dx

Let:f(x)=1+x4dxDifferentiate and equal to 0 to find extrema:f′(x)=12(1+x4)−12⋅4x3=2x31+x40=2x31+x4⇒x=0f(0)=1+04=1(0,1) is absolute minimum. Check the edge cases:f(−1)=2=f(1)⇒m=1,M=2Estimate the value of the integral:m(b−a)≤∫abf(x)dx≤M(b−a)1(1−(−1))≤∫−111+x4≤2(1−(−1))2≤∫−111+x4≤22

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OlmekinjP

Answered question

2021-10-09

Use Property 8 to estimate the value of the integral.

\int_{- 1}^{1} \sqrt{1 + x^{4}} d x

Answer & Explanation

rogreenhoxa8

Skilled2021-10-10Added 109 answers

Let:

f (x) = \sqrt{1 + x^{4}} d x

Differentiate and equal to 0 to find extrema:

f^{'} (x) = \frac{1}{2} {(1 + x^{4})}^{\frac{- 1}{2}} \cdot 4 x^{3}

= \frac{2 x^{3}}{\sqrt{1 + x^{4}}}

0 = \frac{2 x^{3}}{\sqrt{1 + x^{4}}}

\Rightarrow x = 0

f (0) = \sqrt{1 + 0^{4}} = 1

(0,1) is absolute minimum. Check the edge cases:

f (- 1) = \sqrt{2} = f (1)

\Rightarrow m = 1, M = \sqrt{2}

Estimate the value of the integral:

m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)

1 (1 - (- 1)) \leq \int_{- 1}^{1} \sqrt{1 + x^{4}} \leq \sqrt{2} (1 - (- 1))

2 \leq \int_{- 1}^{1} \sqrt{1 + x^{4}} \leq 2 \sqrt{2}

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