Irrerbthist6n

2021-12-14

Find formulas for the functions represented by the integrals.
${\int }_{1}^{{x}^{2}}tdt$

Ronnie Schechter

Step 1: To determine
Find formula for the function represented by the given integral:
${\int }_{1}^{{x}^{2}}tdt$
Step 2:Formula used
$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$ where C is the constant of integration
Step 3:Solution
Consider the given integral:
${\int }_{1}^{{x}^{2}}tdt$
$=\frac{{t}^{2}}{2}{\mid }_{1}^{{x}^{2}}$
$=\frac{1}{2}\left({\left({x}^{2}\right)}^{2}-{1}^{2}\right)$
$=\frac{1}{2}\left({x}^{4}-1\right)$
Hence, the function represented by the given integral is $\frac{1}{2}\left({x}^{4}-1\right)$
Step 4:Conclusion
Hence, the function represented by the given integral is $\frac{1}{2}\left({x}^{4}-1\right)$

Shannon Hodgkinson

${\int }_{1}^{{x}^{2}}tdt$
Evaluate the indefinite integral
$\int tdt$
Evaluate the integral
$\frac{{t}^{2}}{2}$
Return the limits
$\frac{{t}^{2}}{2}{\mid }_{1}^{{x}^{2}}$
Calculate the expression
$\frac{{\left({x}^{2}\right)}^{2}}{2}-\frac{{1}^{2}}{2}$
Simplify
Solution
$\frac{{x}^{4}-1}{2}$

nick1337

It is required to calculate:
$\int tdt$
Integral of a power function:

$=\frac{{t}^{2}}{2}$
Problem solved:
$\int tdt$
$=\frac{{t}^{2}}{2}+C$

Do you have a similar question?