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

Solve the integral.
$$\displaystyle{\int_{{{1}}}^{{{9}}}}{t}^{{-{\frac{{{1}}}{{{2}}}}}}{\left.{d}{t}\right.}$$

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Drood1980
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.
###### Have a similar question?
Kathleen Ashton
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