Question
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\( \int_{0}\left(-x^{2}+16 x\right) d x+\int_{8}(-8 x+128) d x \) (b) What is the area of the region in terms of \( y \) ?

Ask by Brewer Williams. in the United States
Mar 16,2025

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The area of the region is \( \frac{1280}{3} \).

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The Deep Dive

To solve the provided integrals, we begin by calculating each integral separately. First, evaluate: \[ \int_{0} (-x^2 + 16x) \, dx. \] The antiderivative is: \[ -\frac{x^3}{3} + 8x^2. \] Now, we compute the definite integral from 0 to a point \( a \): \[ \left[-\frac{x^3}{3} + 8x^2\right]_{0}^{a} = \left(-\frac{a^3}{3} + 8a^2\right) - 0 = -\frac{a^3}{3} + 8a^2. \] Next, evaluate: \[ \int_{8} (-8x + 128) \, dx. \] The antiderivative is: \[ -4x^2 + 128x. \] Now, we compute the definite integral from \( 8 \) to \( b \): \[ \left[-4x^2 + 128x\right]_{8}^{b} = \left(-4b^2 + 128b\right) - \left(-4(8^2) + 128(8)\right). \] Calculating the second part gives: \[ -4(64) + 1024 = -256 + 1024 = 768. \] Now, combining both integrals, we'll evaluate at specific endpoints or combine the expressions if you are asking for area in terms of \( y \). To express the area in terms of \( y \), we need to understand the regions under the curves and between the intersections, which can also be framed with respect to geometry or algebraic manipulation typically leading to a closed-form solution. If you need further details on integration limits in a visual context, or a specific intersection of \( y \) to these curves, let us know!

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