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homework9: problem 7 (1 point) calculate the following antiderivatives:…

Question

homework9: problem 7
(1 point)
calculate the following antiderivatives:
(a) $\int x^{5} d x=\square + c$
(b) $\int x^{6 / 5} d x=\square + c$
(c) $\int x^{-4} \sqrt{x} d x=\square + c$
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Explanation:

Part (a)

Step1: Apply power rule for integration

The power rule for integration states that $\int x^n dx = \frac{x^{n + 1}}{n + 1} + C$ (where $n
eq - 1$). For $\int x^{5}dx$, here $n = 5$.
$$\int x^{5}dx=\frac{x^{5 + 1}}{5+1}+C$$

Step2: Simplify the expression

Simplify the exponent and the denominator: $\frac{x^{6}}{6}+C$

Step1: Apply power rule for integration

For $\int x^{\frac{6}{5}}dx$, using the power rule $\int x^n dx=\frac{x^{n + 1}}{n + 1}+C$ with $n=\frac{6}{5}$.
$$\int x^{\frac{6}{5}}dx=\frac{x^{\frac{6}{5}+1}}{\frac{6}{5}+1}+C$$

Step2: Simplify the exponent and denominator

First, calculate the exponent: $\frac{6}{5}+1=\frac{6 + 5}{5}=\frac{11}{5}$. Then the denominator: $\frac{6}{5}+1=\frac{11}{5}$, so the reciprocal is $\frac{5}{11}$. So we have $\frac{5}{11}x^{\frac{11}{5}}+C$

Step1: Rewrite the integrand

Rewrite $x^{-4}\sqrt{x}$ as $x^{-4}\cdot x^{\frac{1}{2}}$. Using the rule $a^m\cdot a^n=a^{m + n}$, we get $x^{-4+\frac{1}{2}}=x^{-\frac{8 - 1}{2}}=x^{-\frac{7}{2}}$

Step2: Apply power rule for integration

Using $\int x^n dx=\frac{x^{n+1}}{n + 1}+C$ with $n =-\frac{7}{2}$.
$$\int x^{-\frac{7}{2}}dx=\frac{x^{-\frac{7}{2}+1}}{-\frac{7}{2}+1}+C$$

Step3: Simplify the exponent and denominator

Calculate the exponent: $-\frac{7}{2}+1=-\frac{7 - 2}{2}=-\frac{5}{2}$. The denominator: $-\frac{7}{2}+1=-\frac{5}{2}$, so the reciprocal is $-\frac{2}{5}$. Thus, we have $-\frac{2}{5}x^{-\frac{5}{2}}+C$ (or equivalently $-\frac{2}{5x^{\frac{5}{2}}}+C$)

Answer:

$\frac{x^{6}}{6}$

Part (b)