Evaluate the definite integral.\int_0^1\sqrt[3]{1+7x}dx

iricky827b 2021-11-15 Answered

Evaluate the definite integral.
\(\int_0^1\sqrt[3]{1+7x}dx\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Huses1969
Answered 2021-11-16 Author has 7917 answers

Compute the definite integral
\(\int_0^1\sqrt[3]{1+7x}dx\)
For the integrand \(\sqrt[3]{7x+1}\), substitute \(\displaystyle{u}={7}{x}+{1}\) and \(\displaystyle{d}{u}={7}{\left.{d}{x}\right.}\)
This gives a new lower bound
\(\displaystyle{u}={7}\times{0}+{1}={1}\)
and upper bound \(\displaystyle{u}={7}\times{1}+{1}={8}\)
\(\displaystyle={\frac{{{1}}}{{{7}}}}{\int_{{1}}^{{8}}}{3}\sqrt{{{u}}}{d}{u}\)
Apply the fundamental theorem of calculus
The antiderivative of \(\sqrt[3]{u}\) is \(\displaystyle{\frac{{{3}{u}}}{{{4}}}}^{{\frac{{4}}{{3}}}}\)
\(\displaystyle{\left\lbrace\int{u}^{{n}}{d}{u}={\frac{{{u}^{{{n}+{1}}}}}{{{n}+{1}}}}\right\rbrace}\)
\(\displaystyle={{\left[{\frac{{{3}{u}^{{\frac{{4}}{{3}}}}}}{{{28}}}}\right]}_{{1}}^{{8}}}\)
Evaluate the antiderivative at the limits and subtract
\(\displaystyle{{\left[{\frac{{{3}{u}^{{\frac{{4}}{{3}}}}}}{{{28}}}}\right]}_{{1}}^{{8}}}={\frac{{{3}\times{8}^{{\frac{{4}}{{3}}}}}}{{{28}}}}-{\frac{{{3}\times{1}^{{\frac{{4}}{{3}}}}}}{{{28}}}}={\frac{{{45}}}{{{28}}}}\)
Therefore,
\(\int_0^1\sqrt[3]{1+7x}dx=\frac{45}{28}\)

Not exactly what you’re looking for?
Ask My Question
0
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question
...