If R=[0,1]x [0,1], show that $0\le \int {\int}_{R}\mathrm{sin}(x+y)dA\le 1$

skilpadw3
2022-08-01
Answered

If R=[0,1]x [0,1], show that $0\le \int {\int}_{R}\mathrm{sin}(x+y)dA\le 1$

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phravincegrln2

Answered 2022-08-02
Author has **19** answers

If R = [0, 1] x [0, 1], show that $0\le \int {\int}_{R}\mathrm{sin}(x+y)dA\le 1$

the region $R=((x,y)|0\le x\le 1,0\le y\le 1)$

therefore the integral becomes.

${\int}_{0}^{1}{\int}_{0}^{1}\mathrm{sin}(x+y)dxdy$

$={\int}_{0}^{1}[-\mathrm{cos}(x+y){]}_{0}^{1}dy$

$=-{\int}_{0}^{1}[\mathrm{cos}(1+y)-\mathrm{cos}(y)]dy$

$=-[\mathrm{sin}(1+y)-\mathrm{sin}(y){]}_{0}^{1}$

$=-[(\mathrm{sin}2-\mathrm{sin}1)-(\mathrm{sin}1-\mathrm{sin}0)]$

$=2\mathrm{sin}1-\mathrm{sin}2\cong 0.774$

$0\le 0.774\le 1$

the region $R=((x,y)|0\le x\le 1,0\le y\le 1)$

therefore the integral becomes.

${\int}_{0}^{1}{\int}_{0}^{1}\mathrm{sin}(x+y)dxdy$

$={\int}_{0}^{1}[-\mathrm{cos}(x+y){]}_{0}^{1}dy$

$=-{\int}_{0}^{1}[\mathrm{cos}(1+y)-\mathrm{cos}(y)]dy$

$=-[\mathrm{sin}(1+y)-\mathrm{sin}(y){]}_{0}^{1}$

$=-[(\mathrm{sin}2-\mathrm{sin}1)-(\mathrm{sin}1-\mathrm{sin}0)]$

$=2\mathrm{sin}1-\mathrm{sin}2\cong 0.774$

$0\le 0.774\le 1$

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A car braked with a constant deceleration of $40\text{ft}/{\text{s}}^{2}$, producing skid marks measuring 160 ft before coming to a stop. How fast was the car travelling when the brakes were first applied?

I know I can solve this problem using kinematics equations from physics; using ${v}_{f}^{2}={v}_{i}^{2}+2ad$ yields an initial velocity of 113 ft/s. However, I am supposed to be using antiderivatives and not physics. So far, I figured that if $a(t)=-40$ then $v(t)=-40t+{c}_{1}$ and $d(t)=-20{t}^{2}+{c}_{1}x+{c}_{2}$, where ${c}_{1}$ and ${c}_{2}$ are constants. I'm not quite sure what my next step should be... any suggestions?