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Question 1
7. (a) Find \[ \int (2x - 1)^{\frac{3}{2}} \, dx \] giving your answer in its simplest form. (b) Find the value of \( k \). The finite region \( S \), shown shad... show full transcript
Step 1
Answer
To solve the integral ( \int (2x - 1)^{\frac{3}{2}} , dx ), we use substitution. Let ( u = 2x - 1 ), then ( du = 2dx ) or ( dx = \frac{du}{2} ). Now, substituting this in:
[ \int (2x - 1)^{\frac{3}{2}} , dx = \frac{1}{2} \int u^{\frac{3}{2}} , du = \frac{1}{2} \cdot \frac{2}{5} u^{\frac{5}{2}} + C = \frac{1}{5} (2x - 1)^{\frac{5}{2}} + C ]
Thus, the answer is ( \frac{1}{5} (2x - 1)^{\frac{5}{2}} + C ).
Step 2
Answer
We know that the curve ( y = (2x - 1)^{\frac{3}{2}} ) intersects the line ( y = 8 ) at point ( P(k, 8) ). Setting ( 8 = (2k - 1)^{\frac{3}{2}} ), we cube both sides:
[ 512 = 2k - 1 ] [ 2k = 513 ] [ k = \frac{513}{2} = 256.5 ]
Thus, ( k = 256.5 ).
Step 3
Answer
To find the volume generated by rotating the region ( S ) about the x-axis, we use the formula:
[ V = \pi \int_{a}^{b} [f(x)]^2 , dx ]
where ( f(x) = (2x-1)^{\frac{3}{2}} ) and the limits are from ( x = \frac{1}{2} ) to ( x = k ). Thus, we need to evaluate:
[ V = \pi \int_{\frac{1}{2}}^{k} (2x-1)^3 , dx ] Calculating, we first find the integral: [ \int (2x-1)^3 , dx = \frac{(2x-1)^4}{8} + C ] So, [ V = \pi \left[ \frac{(2x-1)^4}{8} \right]_{\frac{1}{2}}^{k} = \pi \left[ \frac{(2k-1)^4}{8} - \frac{(0)^4}{8} \right] = \frac{\pi (2k-1)^4}{8} ] Substituting the value of ( k ) gives the final volume.
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