Evaluate the definite integral. \int_{0}^{2}\frac{x^{2}-2}{x+1}dx

arneguet9k 2021-11-17 Answered
Evaluate the definite integral.
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{\frac{{{x}^{{{2}}}-{2}}}{{{x}+{1}}}}{\left.{d}{x}\right.}\)

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Expert Answer

Julie Mathew
Answered 2021-11-18 Author has 7282 answers
Step 1
Using the following integrations to solve:
\(\displaystyle\int{x}^{{{n}}}{\left.{d}{x}\right.}={\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}}+{C}\)
\(\displaystyle\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}+{C}\)
where C is the constant when indefinite integral is there, but in question we have definite integral so after solving the integration, given limits have to be substituted.
Step 2
Solving the integration as,
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{\frac{{{x}^{{{2}}}-{2}}}{{{x}+{1}}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{{2}}}}{\left({\left({x}-{1}\right)}-{\left({\frac{{{1}}}{{{x}+{1}}}}\right)}\right)}{\left.{d}{x}\right.}\)
Now integrating it using the formulas,
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{\frac{{{x}^{{{2}}}-{2}}}{{{x}+{1}}}}{\left.{d}{x}\right.}={{\left[{\frac{{{x}^{{{2}}}}}{{{2}}}}-{x}-{\ln}{\left|{x}+{1}\right|}\right]}_{{{0}}}^{{{2}}}}={\frac{{{1}}}{{{2}}}}{\left({4}\right)}-{2}-{\ln}{\left|{3}\right|}\)
\(\displaystyle={2}-{2}-{\ln}{\left|{3}\right|}\)
\(\displaystyle=-{\ln}{\left|{3}\right|}\)
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Troy Lesure
Answered 2021-11-19 Author has 8580 answers
Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
\(\displaystyle{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}\)
Step 2: In this case, \(\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{2}}}-{2}}}{{{x}+{1}}}}\). Find its integral.
\(\displaystyle{\frac{{{x}^{{{2}}}}}{{{2}}}}-{x}-{\ln{{\left({x}+{1}\right)}}}{{\mid}_{{{0}}}^{{{2}}}}\)
Step 3: Since \(\displaystyle{F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}\), expand the above into F(2)−F(0):
\(\displaystyle{\left({\frac{{{2}^{{{2}}}}}{{{2}}}}-{2}-{\ln{{\left({2}+{1}\right)}}}\right)}-{\left({\frac{{{0}^{{{2}}}}}{{{2}}}}-{0}-{\ln{{\left({0}+{1}\right)}}}\right)}\)
Step 4: Simplify.
\(\displaystyle-{\ln{{3}}}\)
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