Explanation Given: ∫ x 1 x 2 ( x − x 1 ) ( x − x 2 ) d x = 1 6 ( x 1 − x 2 ) 3 V = π ∫ x 1 x 1 f ( x ) d x . Let x 1 and x 2 be the roots of f ( x ) = a x + b − ( m x + c ) 2 , where x 1 < x 2 Calculation: We need to verify the formula ∫ x 1 x 2 ( x − x 1 ) ( x − x 2 ) d x = 1 6 ( x 1 − x 2 ) 3 . ∫ x 1 x 2 ( x − x 1 ) ( x − x 2 ) d x = ∫ x 1 x 2 ( x 2 − x ( x 1 + x 2 ) + x 1 x 2 ) d x = [ x 3 3 − x 2 2 ( x 1 + x 2 ) + x 1 x 2 ( x ) ] x 2 x 1 = ( x 2 3 − x 1 3 ) 3 − ( x 2 2 − x 1 2 ) 2 ( x 1 + x 2 ) + ( x 1 x 2 ) ( x 2 − x 1 ) = 2 x 2 3 − 2 x 1 3 − 3 x 2 2 x 1 + 3 x 1 3 − 3 x 2 3 + 3 x 1 2 x 2 + 6 x 1 x 2 2 − 6 x 1 2 x 2 6 = − x 2 3 + x 1 3 + 3 x 2 2 x 1 − 3 x 1 2 x 2 6 = ( x 1 − x 2 ) 3 6 ... ( 1 ) Thus, the given formula is verified. Now, we need to find out the volume of the solid obtained by rotating the shaded region about the x -axis. From the given figure we can see that the shaded region is bounded by the curves, y = a x + b and y = m x + c . Assume, f ( x ) = a x + b and g ( x ) = m x + c . Now, we have to find out the point of intersection of the two given curves and we get: f ( x ) = g ( x ) a x + b = m x + c a x + b = ( m x + c ) 2 m 2 x 2 + ( 2 m c − a ) x + c 2 − b = 0 ... ( 2 ) Now, assume f ( x ) = m 2 x 2 + ( 2 m c − a ) x + c 2 − b = 0 ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 6.3, Problem 65E

To determine

We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

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Chapter 6.3, Problem 64E

Chapter 6.3, Problem 66E

Chapter 6 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 6.1 - Prob. 1PQ Ch. 6.1 - Prob. 2PQ Ch. 6.1 - Prob. 3PQ Ch. 6.1 - Prob. 4PQ Ch. 6.1 - Prob. 5PQ Ch. 6.1 - Prob. 6PQ Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E

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We need to verify the formula ∫ x 1 x 2 ( x − x 1 ) ( x − x 2 ) d x = 1 6 ( x 1 − x 2 ) 3 . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = π 6 B H 2 .

Solution Summary: The author proves that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = 6 B H 2.

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We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

Want to see the full answer?

Chapter 6 Solutions

We need to verify the formula ∫ x 1 x 2 ( x − x 1 ) ( x − x 2 ) d x = 1 6 ( x 1 − x 2 ) 3 . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = π 6 B H 2 .

Solution Summary: The author proves that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = 6 B H 2.

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We need to verify the formula ∫x1x2(x−x1)(x−x2)dx=16(x1−x2)3. Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V=π6BH2.

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Chapter 6 Solutions

We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .