How about solution. ( I = ?)

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2 Answers
Jarob R G.
Apr 28, 2018

I=−104e−2sin(2)

Explanation:

We wish to know what the following integral evaluates to:

I=∫20∫10−104cos(2x)e−2dxdz

Start by removing the constants from the integrand.

I=−104e−2∫20∫10cos(2x)dxdz

We will let C=−104e−2 so that our integral is visually easier to work with.

Integrate cos(2x) with respect to x and evaluate from 0 to 1.

I=C∫20[12sin(2x)]10dz=C2∫20sin(2)dz

Integrate the (constant) term sin(2) with respect to z and evaluate from 0 to 2.

I=C2[sin(2)z]20=C22sin(2)=Csin(2)

Finally, substitute our original constants back into C.

I=−104e−2sin(2)

This is our final answer.

A08
Apr 28, 2018

Given

I=∫20∫10−104cos(2x)e−2dxdz
⇒I=−104e−2∫20∫10cos(2x)dxdz

First Integrate outer integral with respect to z

I=−104e−2∣∣ ∣∣(∫10cos(2x)dx)z∣∣ ∣∣20
⇒I=−104e−2×2∫10cos(2x)dx

Now Integrate with respect to x

I=−104e−2×2∣∣∣12sin(2x)∣∣∣10
⇒I=−104e−2sin2

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