"Find indefinite integral. ∫t2dt" was answered by our experts

High School
Calculus and Analysis
Calculus 2
Applications of integrals

reproacht3

Answered question

2021-12-19

Find indefinite integral.

\int \frac{\sqrt{t}}{2} d t

Answer & Explanation

Steve Hirano

Beginner2021-12-20Added 34 answers

Step 1
Consider the indefinite integral

\int \frac{\sqrt{t}}{2} d t

In order to evaluate this indefinite integral, use the power rule of integration

\int x^{n} d x = \frac{x^{n + 1}}{n + 1}

Step 2
Using the exponent property

\sqrt{n} {a} = a^{\frac{1}{n}}

, the given integral can be written as

\int \frac{\sqrt{t}}{2} d t = \frac{1}{2} \int \sqrt{t} d t

= \frac{1}{2} \int t^{\frac{1}{2}} d t

Now using the power rule of integration, we get

\int \frac{\sqrt{t}}{2} d t = \frac{1}{2} \int t^{\frac{1}{2}} d t

= \frac{1}{2} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C

= \frac{1}{2} [\frac{t^{\frac{1 + 2}{2}}}{\frac{1 + 2}{2}}] + C

= \frac{1}{2} [\frac{t^{\frac{3}{2}}}{\frac{3}{2}}] + C

= \frac{1}{2} [\frac{2 t^{\frac{3}{2}}}{3}] + C

= \frac{t^{\frac{3}{2}}}{3} + C

Thus, the indefinite integral is

\int \frac{\sqrt{t}}{2} d t = \frac{t^{\frac{3}{2}}}{3} + C

\int \frac{\sqrt{t}}{2} d t = \frac{t \sqrt{t}}{3} + C

Here C is the integration constant.

levurdondishav4

Beginner2021-12-21Added 38 answers

Given:
It is required to calculate:

\int \frac{\sqrt{t}}{2} d t

Lets

nick1337

Expert2021-12-28Added 777 answers

$\int \frac{\sqrt{t}}{2} d t$
$\frac{1}{2} \times \int \sqrt{t} d t$
$\frac{1}{2} \times \int t^{\frac{1}{2}} d t$
$\frac{1}{2} \times \frac{2 t \sqrt{t}}{3}$
$\frac{t \sqrt{t}}{3}$
Answer:
$\frac{t \sqrt{t}}{3} + C$

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