grislingatb

2021-11-23

Solve the integral.
${\int }_{1}^{9}{t}^{-\frac{1}{2}}dt$

Drood1980

Step 1
To find:
The definite integral of ${\int }_{1}^{9}{t}^{-\frac{1}{2}}dt$.
Formula used:
Power rule of integration:
${\left({x}^{n}\right)}^{\prime }=n{x}^{n-1}$
Calculation:
The definite integral of ${\int }_{1}^{9}{t}^{-\frac{1}{2}}dt$ can be obtained as,
${\int }_{1}^{9}{t}^{-\frac{1}{2}}dt={\left[\frac{{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]}_{1}^{9}$
$=2{\left[\sqrt{t}\right]}_{1}^{9}$
$=2\left[\sqrt{9}-\sqrt{1}\right]$
=4
Step 2
Thus, the integral of ${\int }_{1}^{9}{t}^{-\frac{1}{2}}dt$ is 4.

Kathleen Ashton

Step 1: Use Negative Power Rule: ${x}^{-a}=\frac{1}{{x}^{a}}$.
${\int }_{1}^{9}\frac{1}{\sqrt{t}}dt$
Step 2: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
${\int }_{a}^{b}f\left(x\right)dx=F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$
Step 3: In this case, $f\left(t\right)=\frac{1}{\sqrt{t}}$. Find its integral.
$2\sqrt{t}{\mid }_{1}^{9}$
Step 4: Since $F\left(t\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$, expand the above into F(9)−F(1):
$2\sqrt{9}-2\sqrt{1}$
Step 5: Simplify.
4

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