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\( \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x-4) d x=\square \) (Type an exact answer.)

Ask by Wade Lambert. in the United States
Mar 14,2025

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Answer

The integral evaluates to \( 2 - 36\pi \).

Solution

Calculate the integral \( \int_{-9\pi/2}^{9\pi/2}(\cos(x)-4)dx \). Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{\frac{-9\pi }{2}}^{\frac{9\pi }{2}} \left(\cos\left(x\right)-4\right) dx\) - step1: Rewrite the fraction: \(\int_{-\frac{9\pi }{2}}^{\frac{9\pi }{2}} \left(\cos\left(x\right)-4\right) dx\) - step2: Evaluate the integral: \(\int \left(\cos\left(x\right)-4\right) dx\) - step3: Use properties of integrals: \(\int \cos\left(x\right) dx+\int -4 dx\) - step4: Evaluate the integral: \(\sin\left(x\right)+\int -4 dx\) - step5: Evaluate the integral: \(\sin\left(x\right)-4x\) - step6: Return the limits: \(\left(\sin\left(x\right)-4x\right)\bigg |_{-\frac{9\pi }{2}}^{\frac{9\pi }{2}}\) - step7: Calculate the value: \(2-36\pi \) The integral of \( \cos(x) - 4 \) from \( -\frac{9\pi}{2} \) to \( \frac{9\pi}{2} \) is \( 2 - 36\pi \).

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To solve the integral \( \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x-4) d x \), we can split this into two parts: 1. The integral of \( \cos x \): \[ \int_{-9 \pi / 2}^{9 \pi / 2} \cos x \, dx = [\sin x]_{-9 \pi / 2}^{9 \pi / 2} \] Notice that \( \sin(-9 \pi / 2) = \sin(-\frac{9\pi}{2}) = 1 \) and \( \sin(9 \pi / 2) = \sin(\frac{9\pi}{2}) = -1 \). Therefore, \[ \sin(9 \pi / 2) - \sin(-9 \pi / 2) = -1 - 1 = -2. \] 2. The integral of \(-4\): \[ \int_{-9 \pi / 2}^{9 \pi / 2} -4 \, dx = -4 \left( x \right) \Big|_{-9 \pi / 2}^{9 \pi / 2} = -4 \left(9 \pi / 2 - (-9 \pi / 2)\right) = -4 (9 \pi) = -36 \pi. \] Now, combine both results: \[ \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x - 4) \, dx = (-2) + (-36\pi) = -36\pi - 2. \] Thus, the final answer is: \[ \boxed{-36\pi - 2}. \]

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