Get the solution to "Evaluate the integral. ∫14t2+1tdt"

Marvin Mccormick

Answered question

2021-10-12

Evaluate the integral.

\int_{1}^{4} \frac{t^{2} + 1}{\sqrt{t}} d t

saiyansruleA

Skilled2021-10-13Added 110 answers

Step 1
Consider the integrals:

\int_{1}^{4} \frac{t^{2} + 1}{\sqrt{t}} d t

Step 2
To solve the given integrals:

\int_{1}^{4} \frac{t^{2} + 1}{\sqrt{t}} d t = \int_{1}^{4} \frac{t^{2}}{\sqrt{t}} d t + \int_{1}^{4} \frac{1}{\sqrt{t}} d t

\int_{1}^{4} t^{2 - \frac{1}{2}} d t + \int_{1}^{4} t^{- \frac{1}{2}} d t

= {[\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}]}_{1}^{4} + {[\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}]}_{1}^{4}

= \frac{2}{5} [{(4)}^{\frac{5}{2}} - {(1)}^{\frac{5}{2}}] + 2 [{(4)}^{\frac{1}{2}} - {(1)}^{\frac{1}{2}}]

further solving it,

= \frac{2}{5} (2^{5} - 1) + 2 (2 - 1)

= \frac{62}{5} + 2

= \frac{62 + 10}{5}

= \frac{72}{5}

So, the value of integral is

\frac{72}{5}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?