How about solution. ( I = ?)

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2 Answers
Jarob R G.
Apr 28, 2018

I = -10^4 e^(-2) sin(2)I=−104e−2sin(2)

Explanation:

We wish to know what the following integral evaluates to:

I = int_0^2 int_0^1 -10^4 cos(2x) e^(-2) dx dzI=∫20∫10−104cos(2x)e−2dxdz

Start by removing the constants from the integrand.

I = -10^4 e^(-2) int_0^2 int_0^1 cos(2x) dx dzI=−104e−2∫20∫10cos(2x)dxdz

We will let C = -10^4 e^(-2)C=−104e−2 so that our integral is visually easier to work with.

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

I = C int_0^2 [1/2 sin(2x)]_0^1 dz = C/2 int_0^2 sin(2) dzI=C∫20[12sin(2x)]10dz=C2∫20sin(2)dz

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

I = C/2 [sin(2)z]_0^2 = C/2 2sin(2) = Csin(2)I=C2[sin(2)z]20=C22sin(2)=Csin(2)

Finally, substitute our original constants back into CC.

I = -10^4 e^(-2) sin(2)I=−104e−2sin(2)

This is our final answer.

A08
Apr 28, 2018

Given

I = int_0^2 int_0^1 -10^4 cos(2x) e^(-2) dx dzI=∫20∫10−104cos(2x)e−2dxdz
=>I = -10^4 e^(-2) int_0^2 int_0^1 cos(2x) dx dz⇒I=−104e−2∫20∫10cos(2x)dxdz

First Integrate outer integral with respect to zz

I = -10^4 e^(-2) | (int_0^1 cos(2x) dx)z|_0^2 I=−104e−2∣∣ ∣∣(∫10cos(2x)dx)z∣∣ ∣∣20
=>I = -10^4 e^(-2)xx2 int_0^1 cos(2x) dx ⇒I=−104e−2×2∫10cos(2x)dx

Now Integrate with respect to xx

I = -10^4 e^(-2)xx2 | 1/2 sin(2x) |_0^1 I=−104e−2×2∣∣∣12sin(2x)∣∣∣10
=>I = -10^4 e^(-2) sin2⇒I=−104e−2sin2

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