Expert Answer to "Calculate the iterated integral. ∫05(∫14(2xy4+3)dy)dx"

smileycellist2

Answered question

2021-11-08

Calculate the iterated integral.

\int_{0}^{5} (\int_{1}^{4} (2 x y^{4} + 3) d y) d x

mhalmantus

Skilled2021-11-09Added 105 answers

Step 1
To calculate the iterated integral

\int_{0}^{5} (\int_{1}^{4} (2 x y^{4} + 3) d y) d x

Step 2
First we integrate the inside integral with respect to y and then integrate with respect to x,

(\int_{1}^{4} (2 x y^{4} + 3) d y) = \int_{1}^{4} 2 x y^{4} d y + \int_{1}^{4} 3 d y

= 2 x \int_{1}^{4} y^{4} d y + 3 \int_{1}^{4} d y

= 2 x {(\frac{y^{5}}{5})}_{1}^{4} + 3 {(y)}_{1}^{4}

= 2 x (\frac{1024}{5} - \frac{1}{5}) + 3 (4 - 1)

= \frac{2046}{5} x + 9

Step 3

\int_{0}^{5} (\int_{1}^{4} (2 x y^{4} + 3) d y) d x = \int_{0}^{5} (\frac{2046}{5} x + 9) d x

= \int_{0}^{5} \frac{2046}{5} x d x + \int_{0}^{5} 9 d x

= \frac{2046}{5} \int_{0}^{5} x d x + 9 \int_{0}^{5} d x

= \frac{2046}{5} {(\frac{x^{2}}{2})}_{0}^{5} + 9 {(x)}_{0}^{5}

= \frac{2046}{5} (\frac{25}{2} - 0) + 9 (5 - 0)

=5115+45
=5160

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