"Calculate the iterated integral. ∫04∫0122ex+3ydx dy" was answered by our community

melodykap

Answered question

2021-01-04

Calculate the iterated integral.

\int_{0}^{4} \int_{0}^{12} 2 e^{x + 3 y} d x d y

SabadisO

Skilled2021-01-05Added 108 answers

Step 1

I = \int_{0}^{4} \int_{0}^{12} 2 e^{x + 3 y} d x d y

Consider the inner integral, name it as

I_{1}

I_{1} = \int_{0}^{12} 2 e^{(x + 3 y)} d x

Take the constant 2 out of the integral sign.

I_{1} = 2 \int_{0}^{12} e^{(x + 3 y)} d x

Step 2
Use substitution,
Substitute t=x+3y

\Rightarrow d t = d x

Change the limits as per the new variable,
When

x \to 0, t \to 3 y

When

x \to 12, t \to 12 + 3 y

Rewrite and solve the new integral as shown,

I_{1} = 2 \int_{3 y}^{12 + 3 y} e^{t} d t

= 2 {(e^{t}]}_{3 y}^{12 + 3 y}

= 2 (e^{12 + 3 y} - e^{3 y})

Step 3
Use the value of

I_{1}

and solve the outer integral.

I = 2 \int_{0}^{4} e^{12 + 3 y} d y - 2 \int_{0}^{4} e^{3 y} d y

I = 2 e^{12} \int_{0}^{4} e^{3 y} d y - 2 \int_{0}^{4} e^{3 y} d y

Evaluate the integrals,

I = \frac{2}{3} e^{12} {(e^{3 y}]}_{0}^{4} - \frac{2}{3} {(e^{3 y}]}_{0}^{4}

I = \frac{2}{3} e^{12} (e^{12} - e^{0}) - \frac{2}{3} (e^{12} - e^{0})

I = \frac{2}{3} e^{12} (e^{12} - 1) - \frac{2}{3} (e^{12} - 1)

Step 4
Combine the terms,

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