Solve the integral. \int_{1}^{9}t^{-\frac{1}{2}}dt

grislingatb 2021-11-23 Answered
Solve the integral.
\(\displaystyle{\int_{{{1}}}^{{{9}}}}{t}^{{-{\frac{{{1}}}{{{2}}}}}}{\left.{d}{t}\right.}\)

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Expert Answer

Drood1980
Answered 2021-11-24 Author has 1770 answers
Step 1
To find:
The definite integral of \(\displaystyle{\int_{{{1}}}^{{{9}}}}{t}^{{-{\frac{{{1}}}{{{2}}}}}}{\left.{d}{t}\right.}\).
Formula used:
Power rule of integration:
\(\displaystyle{\left({x}^{{{n}}}\right)}'={n}{x}^{{{n}-{1}}}\)
Calculation:
The definite integral of \(\displaystyle{\int_{{{1}}}^{{{9}}}}{t}^{{-{\frac{{{1}}}{{{2}}}}}}{\left.{d}{t}\right.}\) can be obtained as,
\(\displaystyle{\int_{{{1}}}^{{{9}}}}{t}^{{-{\frac{{{1}}}{{{2}}}}}}{\left.{d}{t}\right.}={{\left[{\frac{{{t}^{{-{\frac{{{1}}}{{{2}}}}+{1}}}}}{{-{\frac{{{1}}}{{{2}}}}+{1}}}}\right]}_{{{1}}}^{{{9}}}}\)
\(\displaystyle={2}{{\left[\sqrt{{{t}}}\right]}_{{{1}}}^{{{9}}}}\)
\(\displaystyle={2}{\left[\sqrt{{{9}}}-\sqrt{{{1}}}\right]}\)
=4
Step 2
Thus, the integral of \(\displaystyle{\int_{{{1}}}^{{{9}}}}{t}^{{-{\frac{{{1}}}{{{2}}}}}}{\left.{d}{t}\right.}\) is 4.
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Kathleen Ashton
Answered 2021-11-25 Author has 1440 answers
Step 1: Use Negative Power Rule: \(\displaystyle{x}^{{-{a}}}={\frac{{{1}}}{{{x}^{{{a}}}}}}\).
\(\displaystyle{\int_{{{1}}}^{{{9}}}}{\frac{{{1}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}\)
Step 2: 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 3: In this case, \(\displaystyle{f{{\left({t}\right)}}}={\frac{{{1}}}{{\sqrt{{{t}}}}}}\). Find its integral.
\(\displaystyle{2}\sqrt{{{t}}}{{\mid}_{{{1}}}^{{{9}}}}\)
Step 4: Since \(\displaystyle{F}{\left({t}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}\), expand the above into F(9)−F(1):
\(\displaystyle{2}\sqrt{{{9}}}-{2}\sqrt{{{1}}}\)
Step 5: Simplify.
4
0

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