Evaluate

${\int}_{0}^{1}({x}^{5}+{x}^{4}+{x}^{2})\sqrt{4{x}^{3}+5{x}^{2}+10}\phantom{\rule{thickmathspace}{0ex}}dx$

${\int}_{0}^{1}({x}^{5}+{x}^{4}+{x}^{2})\sqrt{4{x}^{3}+5{x}^{2}+10}\phantom{\rule{thickmathspace}{0ex}}dx$

Elliana Molina
2022-11-20
Answered

Evaluate

${\int}_{0}^{1}({x}^{5}+{x}^{4}+{x}^{2})\sqrt{4{x}^{3}+5{x}^{2}+10}\phantom{\rule{thickmathspace}{0ex}}dx$

${\int}_{0}^{1}({x}^{5}+{x}^{4}+{x}^{2})\sqrt{4{x}^{3}+5{x}^{2}+10}\phantom{\rule{thickmathspace}{0ex}}dx$

You can still ask an expert for help

Izabella Henson

Answered 2022-11-21
Author has **20** answers

Answer:

$\begin{array}{rl}{\int}_{0}^{1}({x}^{5}+{x}^{4}+{x}^{2})\sqrt{4{x}^{3}+5{x}^{2}+10}dx& ={\int}_{0}^{1}({x}^{4}+{x}^{3}+x)\sqrt{4{x}^{5}+5{x}^{4}+10{x}^{2}}dx\\ & =\frac{1}{20}{\int}_{0}^{19}\sqrt{u}du=\frac{{19}^{3/2}}{30}\end{array}$

$\begin{array}{rl}{\int}_{0}^{1}({x}^{5}+{x}^{4}+{x}^{2})\sqrt{4{x}^{3}+5{x}^{2}+10}dx& ={\int}_{0}^{1}({x}^{4}+{x}^{3}+x)\sqrt{4{x}^{5}+5{x}^{4}+10{x}^{2}}dx\\ & =\frac{1}{20}{\int}_{0}^{19}\sqrt{u}du=\frac{{19}^{3/2}}{30}\end{array}$

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