For $f(x)={\int}_{x}^{x+1}\mathrm{sin}({\text{e}}^{t})\text{d}t$

Prove that : ${\text{e}}^{x}|f(x)|\le 2$

Prove that : ${\text{e}}^{x}|f(x)|\le 2$

Aarav Atkins
2022-10-12
Answered

For $f(x)={\int}_{x}^{x+1}\mathrm{sin}({\text{e}}^{t})\text{d}t$

Prove that : ${\text{e}}^{x}|f(x)|\le 2$

Prove that : ${\text{e}}^{x}|f(x)|\le 2$

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zupa1z

Answered 2022-10-13
Author has **20** answers

$f(x)={\int}_{x}^{x+1}\mathrm{sin}({\text{e}}^{t})\text{d}t\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle {e}^{x}f(x)={\int}_{x}^{x+1}{e}^{x}\mathrm{sin}({\text{e}}^{t})\text{d}t\le {\int}_{x}^{x+1}{e}^{t}\mathrm{sin}({\text{e}}^{t})\text{d}t}$

Now substitute $z={e}^{t}$ which gives

${e}^{x}f(x)\le {\int}_{{e}^{x}}^{{e}^{x+1}}\mathrm{sin}(\text{z})\text{d}z=\mathrm{cos}({e}^{x})-\mathrm{cos}({e}^{x+1})\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{e}^{x}|f(x)|\le |\mathrm{cos}({e}^{x})|+|\mathrm{cos}({e}^{x+1})|\le 2$

Now substitute $z={e}^{t}$ which gives

${e}^{x}f(x)\le {\int}_{{e}^{x}}^{{e}^{x+1}}\mathrm{sin}(\text{z})\text{d}z=\mathrm{cos}({e}^{x})-\mathrm{cos}({e}^{x+1})\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{e}^{x}|f(x)|\le |\mathrm{cos}({e}^{x})|+|\mathrm{cos}({e}^{x+1})|\le 2$

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