What is the index in the expression 2 2?

Mathematics

2 answers:

Lelechka [254]3 years ago

7 0

Answer:

Step-by-step explanation:

Ostrovityanka [42]3 years ago

7 0

Answer: 2

Step-by-step explanation

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4 years ago

What is 10 - 0.3 (a repeating decimal.)

BabaBlast [244]

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3 years ago

Calculus help please, will give brainiest

Andrew [12]

The most convenient way to capture $D$ is with the parameterization

$D = \left\{(x,y) \mid -1 \le y \le 3 \text{ and } -\dfrac{y+1}2 \le x \le y+1\right\}$

so we only need one iterated integral.

$\displaystyle \iint_D e^{x-y} \, dA = \int_{-1}^3 \int_{-(y+1)/2}^{y+1} e^{x-y} \, dx \, dy$

Compute the integral with respect to $x$ .

$\displaystyle \int_{-(y+1)/2}^{y+1} e^{x-y} \, dx = e^{x-y} \bigg|_{x=-(y+1)/2}^{x=y+1} = e^{(y+1)-y} - e^{-(y+1)/2-y} = e - e^{-(3y+1)/2}$

Compute the remaining integral.

$\displaystyle \int_{-1}^3 \left(e - e^{-(3y+1)/2}\right) \, dy = \left(ey + \frac23 e^{-(3y+1)/2}\right)\bigg|_{y=-1}^{y=3} \\\\ ~~~~~~~~ = \left(3e + \frac23 e^{-(9+1)/2}\right) - \left(-e + \frac23 e^{-(-3+1)/2}\right) \\\\ ~~~~~~~~ = \boxed{\frac{10e}3 + \frac2{3e^5}}$