Integrals Involving Roots — Question 3

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Question 3

Find the geometric area between y=x3y=\sqrt[3]{x} and the xx-axis on [−1,8][-1,8].

Original worksheet page 1: question and worked solution for 1-5-003
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Question 3 – Solution

Step 1: Split at the xx-intercept. The function is negative on [−1,0][-1,0] and positive on [0,8][0,8], so A=−∫−10x1/3dx+∫08x1/3dx.A=-\int_{-1}^0x^{1/3}\,dx+\int_0^8x^{1/3}\,dx. Step 2: Find an antiderivative. ∫x1/3dx=x4/34/3=34x4/3.\int x^{1/3}\,dx=\frac{x^{4/3}}{4/3}=\frac34x^{4/3}. Step 3: Evaluate both pieces. A=−[34x4/3]−10+[34x4/3]08=−(0−34)+(34⋅16−0)=34+12=514.\begin{align*} A&=-\left[\frac34x^{4/3}\right]_{-1}^{0} +\left[\frac34x^{4/3}\right]_{0}^{8}\\ &=-\left(0-\frac34\right)+\left(\frac34\cdot16-0\right)\\ &=\frac34+12=\frac{51}{4}. \end{align*} A=514\boxed{A=\frac{51}{4}}

Original worksheet page 2: question and worked solution for 1-5-003

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