Find the Area Under the Curve y = cube root of x+1 ; [-1,7]

$y = \sqrt[3]{x + 1}$ ; $[- 1, 7]$

Solve by substitution to find the intersection between the curves.

Eliminate the equal sides of each equation and combine.

$\sqrt[3]{x + 1} = 0$

Solve $\sqrt[3]{x + 1} = 0$ for $x$ .

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{x + 1}}^{3} = 0^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x + 1}$ as ${(x + 1)}^{\frac{1}{3}}$ .

${({(x + 1)}^{\frac{1}{3}})}^{3} = 0^{3}$

Simplify the left side.

Simplify ${({(x + 1)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(x + 1)}^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x + 1)}^{\frac{1}{3} \cdot 3} = 0^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x + 1)}^{\frac{1}{3} \cdot 3} = 0^{3}$

Rewrite the expression.

${(x + 1)}^{1} = 0^{3}$

Simplify.

$x + 1 = 0^{3}$

Simplify the right side.

Raising $0$ to any positive power yields $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Substitute $- 1$ for $x$ .

$y = 0$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 1, 0)$

The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region. The regions are determined by the intersection points of the curves. This can be done algebraically or graphically.

$A r e a = \int_{- 1}^{7} \sqrt[3]{x + 1} d x - \int_{- 1}^{7} 0 d x$

Integrate to find the area between $- 1$ and $7$ .

Combine the integrals into a single integral.

$\int_{- 1}^{7} \sqrt[3]{x + 1} - (0) d x$

Subtract $0$ from $\sqrt[3]{x + 1}$ .

$\int_{- 1}^{7} \sqrt[3]{x + 1} d x$

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

Let $u = x + 1$ . Find $\frac{d u}{d x}$ .

Differentiate $x + 1$ .

$\frac{d}{d x} [x + 1]$

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [1]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [1]$

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$1 + 0$

Add $1$ and $0$ .

$1$

Substitute the lower limit in for $x$ in $u = x + 1$ .

$u_{lower} = - 1 + 1$

Add $- 1$ and $1$ .

$u_{lower} = 0$

Substitute the upper limit in for $x$ in $u = x + 1$ .

$u_{upper} = 7 + 1$

Add $7$ and $1$ .

$u_{upper} = 8$

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0 u_{upper} = 8$

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\int_{0}^{8} \sqrt[3]{u} d u$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{u}$ as $u^{\frac{1}{3}}$ .

$\int_{0}^{8} u^{\frac{1}{3}} d u$

By the Power Rule, the integral of $u^{\frac{1}{3}}$ with respect to $u$ is $\frac{3}{4} u^{\frac{4}{3}}$ .

$\frac{3}{4} u^{\frac{4}{3}}]_{0}^{8}$

Substitute and simplify.

Evaluate $\frac{3}{4} u^{\frac{4}{3}}$ at $8$ and at $0$ .

$(\frac{3}{4} \cdot 8^{\frac{4}{3}}) - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Simplify.

Rewrite $8$ as $2^{3}$ .

$\frac{3}{4} \cdot {(2^{3})}^{\frac{4}{3}} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{3}{4} \cdot 2^{3 (\frac{4}{3})} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3}{4} \cdot 2^{3 (\frac{4}{3})} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Rewrite the expression.

$\frac{3}{4} \cdot 2^{4} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Raise $2$ to the power of $4$ .

$\frac{3}{4} \cdot 16 - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Combine $\frac{3}{4}$ and $16$ .

$\frac{3 \cdot 16}{4} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Multiply $3$ by $16$ .

$\frac{48}{4} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Cancel the common factor of $48$ and $4$ .

Factor $4$ out of $48$ .

$\frac{4 \cdot 12}{4} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Cancel the common factors.

Factor $4$ out of $4$ .

$\frac{4 \cdot 12}{4 (1)} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Cancel the common factor.

$\frac{4 \cdot 12}{4 \cdot 1} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Rewrite the expression.

$\frac{12}{1} - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Divide $12$ by $1$ .

$12 - \frac{3}{4} \cdot 0^{\frac{4}{3}}$

Rewrite $0$ as $0^{3}$ .

$12 - \frac{3}{4} \cdot {(0^{3})}^{\frac{4}{3}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$12 - \frac{3}{4} \cdot 0^{3 (\frac{4}{3})}$

Cancel the common factor of $3$ .

Cancel the common factor.

$12 - \frac{3}{4} \cdot 0^{3 (\frac{4}{3})}$

Rewrite the expression.

$12 - \frac{3}{4} \cdot 0^{4}$

Raising $0$ to any positive power yields $0$ .

$12 - \frac{3}{4} \cdot 0$

Multiply $0$ by $- 1$ .

$12 + 0 (\frac{3}{4})$

Multiply $0$ by $\frac{3}{4}$ .

$12 + 0$

Add $12$ and $0$ .

$12$

Find the Area Under the Curve y = cube root of x+1 ; [-1,7]

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