How do you evaluate the definite integral $\int (t^{\frac{1}{3}} - 2)$ $int (t^(1/3)-2)$ from $[- 1, 1]$ $[-1,1]$ ?

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How do you evaluate the definite integral $\int (t^{\frac{1}{3}} - 2)$ $int (t^(1/3)-2)$ from $[- 1, 1]$ $[-1,1]$ ?

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Answer 1 · 2018-03-31T18:30:45

$\int_{- 1}^{1} (t^{\frac{1}{3}} - 2) d t = 4$ $int_-1^1(t^(1/3)-2)dt=4$

Explanation:

$\int_{- 1}^{1} (t^{\frac{1}{3}} - 2) d t = \int_{- 1}^{1} t^{\frac{1}{3}} d t - \int_{- 1}^{1} 2 d t$ $int_-1^1(t^(1/3)-2)dt=int_-1^1t^(1/3)dt-int_-1^1 2dt$

We can split up sums or differences when dealing with integrals.

Now, integrate each term individually and evaluate from $[- 1, 1] .$ $[-1, 1].$

$\frac{3}{4} t^{\frac{4}{3}} {∣ ∣_{- 1}^{1} - 2 d t ∣ ∣}_{- 1}^{1} = (\frac{3}{4} \cdot 1^{\frac{4}{3}} - \frac{3}{4} \cdot - 1^{\frac{4}{3}}) - (2 (1) - 2 (- 1))$ $3/4t^(4/3)|_-1^1-2dt|_-1^1=(3/4*1^(4/3)-3/4*-1^(4/3))-(2(1)-2(-1))$

$1^{\frac{4}{3}} = \sqrt[3]{1^{4}} = 1$ $1^(4/3)=root(3)(1^4)=1$
${(- 1)}^{\frac{4}{3}} = \sqrt[3]{{(- 1)}^{4}} = \sqrt[3]{1} = 1$ $(-1)^(4/3)=root(3)((-1)^4)=root(3)1=1$

So, we then have

$(\frac{3}{4} - \frac{3}{4}) - (2 - (- 2)) = 4$ $(3/4-3/4)-(2-(-2))=4$

$\int_{- 1}^{1} (t^{\frac{1}{3}} - 2) d t = 4$ $int_-1^1(t^(1/3)-2)dt=4$

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How do you evaluate the definite integral $\int (t^{\frac{1}{3}} - 2)$ $int (t^(1/3)-2)$ from $[- 1, 1]$ $[-1,1]$ ?

1 Answer

Explanation:

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How do you evaluate the definite integral $\int (t^{\frac{1}{3}} - 2)$ $int (t^(1/3)-2)$ from $[- 1, 1]$ $[-1,1]$ ?