Calculus Archive: Questions from July 18, 2023
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Calculate the double integral. \[ \iint_{R} \frac{6\left(1+x^{2}\right)}{1+y^{2}} d A, R=\{(x, y) \mid 0 \leq x \leq 3,0 \leq y \leq 1\} \]2 answers -
Calculate the double integral. \[ \iint_{R} \frac{6\left(1+x^{2}\right)}{1+y^{2}} d A, R=\{(x, y) \mid 0 \leq x \leq 3,0 \leq y \leq 1\} \]2 answers -
Evaluate the triple integral. \[ \iiint_{E} y d V \text {, where } E=\{(x, y, z) \mid 0 \leq x \leq 2,0 \leq y \leq x, x-y \leq z \leq x+y\} \]2 answers -
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6. Differentiate the given functions. a) \( y=x^{-4} \) a) \( y=2 \sqrt[4]{x^{3}} \) c) \( y=3 x^{5}-4 x^{3}+9 x-6 \) d) \( f(u)=0.07 u^{4}-1.21 u^{3}+3 u-5.2 \) e) \( f(t)=2 \sqrt{t^{3}}+\frac{4}{\sq2 answers -
7.) Evaluate \( \int_{C} \vec{F} \cdot d \vec{r} \). \[ \begin{array}{l} \vec{F}(x, y, z)=\langle 3 x, 4 y, x z\rangle \\ \vec{r}(t)=\langle 6 t+2,8 t-3,7 t\rangle, 0 \leq t \leq 1 \end{array} \]2 answers -
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integral de V+3/(2V^3-8V) dv inseguro de cómo dividir el denominador cualquier ayuda sería la regla.2 answers
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