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

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

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Huses1969

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,
$$\int_0^1\sqrt[3]{1+7x}dx=\frac{45}{28}$$