Aaron Coleman

2022-04-30

What is $$\int \frac{2}{9}\ufeff\sqrt{4x+5}dx$$?

zavesiljid

Beginner2022-05-01Added 21 answers

Since $$\frac{2}{9}$$ is constant with respect to x, move $$\frac{2}{9}$$ out of the integral.

$$\frac{2}{9}\int \sqrt{4x+5}dx$$

Let u=4x+5. Then du=4dx, so $$\frac{1}{4}du=dx$$. Rewrite using u and du.

$$\frac{2}{9}\int \sqrt{u}\frac{1}{4}du$$

Combine $$\sqrt{u}$$ and $$\frac{1}{4}$$.

$$\frac{2}{9}\int \frac{\sqrt{u}}{4}du$$

Since $$\frac{1}{4}$$ is constant with respect to u, move $$\frac{1}{4}$$ out of the integral.

$$\frac{2}{9}(\frac{1}{4}\int \sqrt{u}du)$$

Simplify the expression.

$$\frac{1}{18}\int {u}^{\frac{1}{2}}du$$

By the Power Rule, the integral of $${u}^{\frac{1}{2}}$$ with respect to u is $$\frac{2}{3}{u}^{\frac{3}{2}}$$

$$\frac{1}{18}(\frac{2}{3}{u}^{\frac{3}{2}}+C)$$

Simplify

$$\frac{1}{27}{u}^{\frac{3}{2}}+C$$

Replace all occurrences of u with 4x+5.

$$\frac{1}{27}(4x+5{)}^{\frac{3}{2}}+C$$

$$\frac{2}{9}\int \sqrt{4x+5}dx$$

Let u=4x+5. Then du=4dx, so $$\frac{1}{4}du=dx$$. Rewrite using u and du.

$$\frac{2}{9}\int \sqrt{u}\frac{1}{4}du$$

Combine $$\sqrt{u}$$ and $$\frac{1}{4}$$.

$$\frac{2}{9}\int \frac{\sqrt{u}}{4}du$$

Since $$\frac{1}{4}$$ is constant with respect to u, move $$\frac{1}{4}$$ out of the integral.

$$\frac{2}{9}(\frac{1}{4}\int \sqrt{u}du)$$

Simplify the expression.

$$\frac{1}{18}\int {u}^{\frac{1}{2}}du$$

By the Power Rule, the integral of $${u}^{\frac{1}{2}}$$ with respect to u is $$\frac{2}{3}{u}^{\frac{3}{2}}$$

$$\frac{1}{18}(\frac{2}{3}{u}^{\frac{3}{2}}+C)$$

Simplify

$$\frac{1}{27}{u}^{\frac{3}{2}}+C$$

Replace all occurrences of u with 4x+5.

$$\frac{1}{27}(4x+5{)}^{\frac{3}{2}}+C$$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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