Question

Is this right? Is my decimal right? Please check my work

∫[0,3] 2√(1 + t^2) dt
I got 24.2984

Answer 1

no. it's still 11.305

I can't check your work, since you didn't show any.

Answer 2

0

∫ 2 √ ( t² + 1 ) dt
3

2 ∫ √ ( t² + 1 ) dt = arsinh ( t ) + t √ ( t² + 1 )

since:

arsinh ( t ) = ln | √ ( t² + 1 ) + t |

then

∫ 2 √ ( t² + 1 ) dt = ln | √ ( t² + 1 ) + t | + t ∙ √ ( t² + 1 ) + C

0
∫ 2 √ ( t² + 1 ) dt = F(3) - F(0)
3

where

F(0) = ln | √ ( 0² + 1 ) + 0 | + 0 ∙ √ ( 0² + 1 ) = ln | √ ( 0 + 1 ) + 0 = ln ( √ 1 ) = ln ( 1 ) = 0

F(3) = ln | √ ( 3² + 1 ) + 3 | + 3 ∙√ ( 3² + 1 ) = ln | √ ( 9 + 1 ) + 3 | + 3 ∙ √ ( 9 + 1 ) =

= ln | 3 + √ 10 | + 3 √ 10 = ln | 3 + √ 10 | + 3 √ 10 = ln ( 6.16227766 ) + 3 ∙ 3.16227766 =

1.818446459 + 9.48683298 = 11.305279439

F(3) - F(0) = 11.305279439 - 0 = 11.305279439

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Remark:

This integration is very difficult to solve.

If you want in google type:

Integration online

when you see list of resus clik on:

w w w. integral-calculator.c o m
Integral Calculator • With Steps

When page be open in rectangle paste:

2 √ ( t² + 1 ) dt

set

Upper bound 3

Lower bound 0

Variable of integration: t

and click

Go!

you will see solution with steps

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