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Question Number 101439 by mhmd last updated on 02/Jul/20

Commented by mhmd last updated on 02/Jul/20

$$helpmesir?$$

Commented by bramlex last updated on 02/Jul/20

$$\left(\mathbb{Q}3\right)\frac{\mathrm{dy}}{{4\mathrm{x}}^2}-\sec 5\mathrm{y}\mathrm{dx}=0$$$$\frac{\mathrm{dy}}{\sec 5\mathrm{y}}={4\mathrm{x}}^2\mathrm{dx}\Rightarrow \int \cos 5\mathrm{y}\mathrm{dy}=\frac{4}{3}{\mathrm{x}}^3+\mathrm{C}$$$$\frac{1}{5}\sin 5\mathrm{y}=\frac{4}{3}{\mathrm{x}}^3+\mathrm{C}★$$

Commented by bemath last updated on 02/Jul/20

$$\left(\mathfrak{q}4\right){8\mathrm{x}}^2=32,\mathrm{x}=\pm 2$$$$\mathcal{T}\mathrm{he}\mathrm{area}=2\underset{0}{\stackrel{2}{\int }}(32-{8\mathrm{x}}^2)\mathrm{dx}$$$$=2{\left[(32\mathrm{x}-\frac{8}{3}{\mathrm{x}}^3)\right]}_0^2$$$$=2(64-\frac{64}{3})=128(1-\frac{1}{3})$$$$=128\times \frac{2}{3}=\frac{256}{3}\mathrm{♠}$$

Commented by mhmd last updated on 02/Jul/20

$$thankyousir$$

Commented by bemath last updated on 04/Jul/20

$$\text{Missing left or extra right}$$$$=\underset{0}{\stackrel{2}{\int }}\left(4\pi \right)\mathrm{dx}={\left[4\pi \mathrm{y}\right]}_0^2$$$$=8\pi ★$$

Commented by mhmd last updated on 02/Jul/20

$$sircanyouhelpmeinquestionone$$

Commented by bobhans last updated on 02/Jul/20

$$\left(\mathrm{Q}5\right)\cos (-2\mathrm{x})=1-\frac{{(-2\mathrm{x})}^2}{2!}+\frac{{(-2\mathrm{x})}^4}{4!}-\frac{{(-2\mathrm{x})}^6}{6!}+...$$

Commented by mhmd last updated on 02/Jul/20

$$sirtheinterval\left(0topi\right)$$

Commented by mhmd last updated on 02/Jul/20

$$canyouhelpmeinquestionone$$

Commented by mhmd last updated on 03/Jul/20

$$nosirtheresult\left(8pi\right)not\left(16pi\right)$$

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