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Answer the following True or False.\newlineSince ddx3x13=x−23\frac{d}{dx}3x^{\frac{1}{3}}=x^{-\frac{2}{3}}, the fundamental theorem of calculus tells us that ∫−11x−23dx=3(113)−3(−1)13=6\int_{-1}^{1}x^{-\frac{2}{3}}dx=3(1^{\frac{1}{3}})-3(-1)^{\frac{1}{3}}=6.\newlineTrue\newlineFalse

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Q. Answer the following True or False.\newlineSince ddx3x13=x−23\frac{d}{dx}3x^{\frac{1}{3}}=x^{-\frac{2}{3}}, the fundamental theorem of calculus tells us that ∫−11x−23dx=3(113)−3(−1)13=6\int_{-1}^{1}x^{-\frac{2}{3}}dx=3(1^{\frac{1}{3}})-3(-1)^{\frac{1}{3}}=6.\newlineTrue\newlineFalse
  1. Verify Derivative: Let's first verify the derivative given in the statement.\newlineGiven the function f(x)=3x13f(x) = 3x^{\frac{1}{3}}, we need to find its derivative f′(x)f'(x).\newlineUsing the power rule for derivatives, we have:\newlinef′(x)=ddx[3x13]=3⋅(13)⋅x(13−1)=x−23f'(x) = \frac{d}{dx} [3x^{\frac{1}{3}}] = 3 \cdot \left(\frac{1}{3}\right) \cdot x^{\left(\frac{1}{3} - 1\right)} = x^{-\frac{2}{3}}.
  2. Apply Fundamental Theorem: Now, let's apply the Fundamental Theorem of Calculus, which states that if FF is an antiderivative of ff on an interval [a,b][a, b], then:\newline∫abf(x) dx=F(b)−F(a)\int_a^b f(x) \, dx = F(b) - F(a).\newlineHere, f(x)=x(−2/3)f(x) = x^{(-2/3)} and F(x)=3x(1/3)F(x) = 3x^{(1/3)} is its antiderivative.
  3. Evaluate Integral: We will now evaluate the integral using the antiderivative: ∫−11x−23 dx=F(1)−F(−1)=3(113)−3((−1)13)\int_{-1}^{1} x^{-\frac{2}{3}} \, dx = F(1) - F(-1) = 3(1^{\frac{1}{3}}) - 3((-1)^{\frac{1}{3}}).
  4. Be Careful with Evaluation: We need to be careful with the evaluation of F(−1)F(-1) because (−1)13(-1)^{\frac{1}{3}} is not equal to −1-1. The cube root of −1-1 is −1-1, so we have: F(1)−F(−1)=3(1)−3(−1)=3+3=6F(1) - F(-1) = 3(1) - 3(-1) = 3 + 3 = 6.
  5. Statement Verification: The statement given is ＂Since (ddx3x(13)=x−(23))(\frac{d}{dx}3x^{(\frac{1}{3})}=x^{-(\frac{2}{3})}), the fundamental theorem of calculus tells us that ∫−11x−(23)dx=3(1(13))−3(−1)(13)=6\int_{-1}^{1}x^{-(\frac{2}{3})}dx=3(1^{(\frac{1}{3})})-3(-1)^{(\frac{1}{3})}=6.＂ Based on our calculations, this statement is true.

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