Question

The Fundamental Theorem of Calculus: Solve each of the following definite integrals.

1. ${\int }_1^{100}\frac{1}{\sqrt[3]{x^4}}dx$.
2. ${\int }_{0.01}^1\frac{1}{\sqrt[3]{x^4}}dx$.
3. ${\int }_1^{100}\frac{1}{\sqrt[4]{x^3}}dx$.
4. ${\int }_{0.01}^1\frac{1}{\sqrt[4]{x^3}}dx$.

Short answer

The solution for the given integrals are,

Part $\left(a\right)$: role="math" localid="1649140409260" ${\int }_1^{100}\frac{1}{\sqrt[3]{x^4}}dx\approx 2.3537$.

Part $\left(b\right)$: ${\int }_{0.01}^1\frac{1}{\sqrt[3]{x^4}}dx\approx 10.9248$.

Part $\left(c\right)$: ${\int }_1^{100}\frac{1}{\sqrt[4]{x^3}}dx\approx 2.1623$.

Part$\left(d\right)$:${\int }_{0.01}^1\frac{1}{\sqrt[4]{x^3}}dx\approx 0.6838$.

Step-by-step solution

Part a Step 1. Given information

${\int }_1^{100}\frac{1}{\sqrt[3]{x^4}}dx$.

Part a Step 2. Now apply the fundamental theorem of calculus in the given integral

$$\begin{gathered} {\int }_1^{100}\frac{1}{\sqrt[3]{x^4}}dx={\int }_1^{100}\frac{1}{{\left(x^4\right)}^{{\displaystyle \frac{1}{3}}}}dx \\ ={\int }_1^{100}{\left(x^4\right)}^{-\frac{1}{3}}dx \\ ={\left[\frac{x^{{\displaystyle \frac{-1}{3}}}}{{\displaystyle \frac{-1}{3}}}\right]}_1^{100} \\ =-3\left[{\left(100\right)}^{\frac{-1}{3}}-1\right] \\ =-3(0.2154-1) \\ \approx 2.3537 \end{gathered}$$

Part b Step 1. Given information

${\int }_{0.01}^1\frac{1}{\sqrt[3]{x^4}}dx$.

Part b Step 2. Now apply the fundamental theorem of calculus in the given integral

$$\begin{gathered} {\int }_{0.01}^1\frac{1}{\sqrt[3]{x^4}}dx=-3{\left[x^{\frac{-1}{3}}\right]}_{0.01}^1 \\ =-3\left[1-4.6416\right] \\ \approx 10.9248 \end{gathered}$$

Part c Step 1. Given information

${\int }_1^{100}\frac{1}{\sqrt[4]{x^3}}dx$.

Part c Now apply the fundamental theorem of calculus in the given interval.

$$\begin{gathered} {\int }_1^{100}\frac{1}{\sqrt[4]{x^3}}dx={\left[x^{\frac{1}{4}}\right]}_1^{100} \\ =\left[{\left({10}^2\right)}^{\frac{1}{4}}-1\right] \\ \approx 2.1623 \end{gathered}$$

Part d Step 1. Given information

${\int }_{0.01}^1\frac{1}{\sqrt[4]{x^3}}dx$.

Part d Step 2. Now apply the fundamental theorem of calculus in the given interval.

$$\begin{gathered} {\int }_{0.01}^1\frac{1}{\sqrt[4]{x^3}}dx={\left[x^{\frac{1}{4}}\right]}_{0.01}^1 \\ =1-{\left(1\times {10}^{-2}\right)}^{\frac{1}{4}} \\ \approx 0.6838 \end{gathered}$$