Problem

Integrate Using u-Substitution integral from 0 to 1 of x square root of 1-x^2 with respect to x

The problem presented is a calculus problem that involves finding the definite integral of the function \(x\sqrt{1-x^2}\) with respect to \(x\) over the interval from 0 to 1 using the method of u-substitution. U-substitution is a technique commonly employed in calculus to simplify the integration process by substituting a part of the integral with a new variable, \(u\), which then simplifies the integration into a more manageable form. The question requires identifying an appropriate substitution, transforming the integral with respect to this new variable \(u\), and then evaluating the definite integral after the substitution has been applied.

$\int_{0}^{1} x \sqrt{1 - x^{2}} d x$

Answer

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Solution:

Step:1

Define $u = 1 - x^{2}$. Consequently, $d u = - 2 x d x$ implies $- \frac{1}{2} d u = x d x$. Transition to $u$ and $d u$ notation.

Step:1.1

Set $u = 1 - x^{2}$ and compute $\frac{d u}{d x}$.

Step:1.1.1

Take the derivative of $1 - x^{2}$: $\frac{d}{d x} [1 - x^{2}]$.

Step:1.1.2

Proceed to differentiate.

Step:1.1.2.1

Using the Sum Rule, the derivative of $1 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x^{2}]$.

Step:1.1.2.2

Since $1$ is a constant, its derivative is $0$: $0 + \frac{d}{d x} [- x^{2}]$.

Step:1.1.3

Calculate $\frac{d}{d x} [- x^{2}]$.

Step:1.1.3.1

The derivative of $- x^{2}$, with the constant $- 1$, is $- \frac{d}{d x} [x^{2}]$: $0 - \frac{d}{d x} [x^{2}]$.

Step:1.1.3.2

Apply the Power Rule, which states $\frac{d}{d x} [x^{n}] = n x^{n - 1}$ for $n = 2$: $0 - (2 x)$.

Step:1.1.3.3

Multiply $2$ by $- 1$: $0 - 2 x$.

Step:1.1.4

Subtract $2 x$ from $0$: $- 2 x$.

Step:1.2

Insert the lower limit into $u = 1 - x^{2}$: $u_{\text{lower}} = 1 - 0^{2}$.

Step:1.3

Simplify the expression.

Step:1.3.1

Simplify each term individually.

Step:1.3.1.1

Any positive power of $0$ is $0$: $u_{\text{lower}} = 1 - 0$.

Step:1.3.1.2

Multiply $- 1$ by $0$: $u_{\text{lower}} = 1 + 0$.

Step:1.3.2

Combine $1$ and $0$: $u_{\text{lower}} = 1$.

Step:1.4

Insert the upper limit into $u = 1 - x^{2}$: $u_{\text{upper}} = 1 - 1^{2}$.

Step:1.5

Simplify the expression.

Step:1.5.1

Simplify each term individually.

Step:1.5.1.1

Any power of one is one: $u_{\text{upper}} = 1 - 1 \cdot 1$.

Step:1.5.1.2

Multiply $- 1$ by $1$: $u_{\text{upper}} = 1 - 1$.

Step:1.5.2

Subtract $1$ from $1$: $u_{\text{upper}} = 0$.

Step:1.6

Utilize the calculated values for $u_{\text{lower}}$ and $u_{\text{upper}}$ to evaluate the definite integral: $u_{\text{lower}} = 1$, $u_{\text{upper}} = 0$.

Step:1.7

Reformulate the integral with $u$, $d u$, and new integration limits: $\int_{1}^{0} \sqrt{u} \frac{1}{- 2} d u$.

Step:2

Streamline the integral.

Step:2.1

Place the negative sign outside the fraction: $\int_{1}^{0} \sqrt{u} \left( - \frac{1}{2} \right) d u$.

Step:2.2

Merge $\sqrt{u}$ and $\frac{1}{2}$: $\int_{1}^{0} - \frac{\sqrt{u}}{2} d u$.

Step:3

Extract the constant $- 1$ from the integral: $- \int_{1}^{0} \frac{\sqrt{u}}{2} d u$.

Step:4

Extract the constant $\frac{1}{2}$ from the integral: $- \left( \frac{1}{2} \int_{1}^{0} \sqrt{u} d u \right)$.

Step:5

Express $\sqrt{u}$ as $u^{\frac{1}{2}}$: $- \frac{1}{2} \int_{1}^{0} u^{\frac{1}{2}} d u$.

Step:6

Integrate $u^{\frac{1}{2}}$ with respect to $u$ using the Power Rule: $- \frac{1}{2} \frac{2}{3} u^{\frac{3}{2}} \Big|_{1}^{0}$.

Step:7

Condense the expression.

Step:7.1

Evaluate $\frac{2}{3} u^{\frac{3}{2}}$ at the limits $0$ and $1$: $- \frac{1}{2} \left( \frac{2}{3} \cdot 0^{\frac{3}{2}} - \frac{2}{3} \cdot 1^{\frac{3}{2}} \right)$.

Step:7.2

Simplify further.

Step:7.2.1

Express $0$ as $0^{2}$: $- \frac{1}{2} \left( \frac{2}{3} \cdot (0^{2})^{\frac{3}{2}} - \frac{2}{3} \cdot 1^{\frac{3}{2}} \right)$.

Step:7.2.2

Apply the rule for exponents $(a^{m})^{n} = a^{m n}$: $- \frac{1}{2} \left( \frac{2}{3} \cdot 0^{2 \cdot \frac{3}{2}} - \frac{2}{3} \cdot 1^{\frac{3}{2}} \right)$.

Step:7.2.3

Eliminate the common factor of $2$.

Step:7.2.3.1

Remove the common factor: $- \frac{1}{2} \left( \frac{2}{3} \cdot 0^{3} - \frac{2}{3} \cdot 1^{\frac{3}{2}} \right)$.

Step:7.2.3.2

Rephrase the expression: $- \frac{1}{2} \left( \frac{2}{3} \cdot 0^{3} - \frac{2}{3} \cdot 1^{\frac{3}{2}} \right)$.

Step:7.2.4

Any positive power of $0$ is $0$: $- \frac{1}{2} \left( \frac{2}{3} \cdot 0 - \frac{2}{3} \cdot 1^{\frac{3}{2}} \right)$.

Step:7.3

Condense the expression.

Step:7.3.1

Multiply $\frac{2}{3}$ by $0$: $- \frac{1}{2} \left( 0 - \frac{2}{3} \cdot 1^{\frac{3}{2}} \right)$.

Step:7.3.2

One raised to any power is one: $- \frac{1}{2} \left( 0 - \frac{2}{3} \cdot 1 \right)$.

Step:7.3.3

Multiply $- 1$ by $1$: $- \frac{1}{2} \left( 0 - \frac{2}{3} \right)$.

Step:7.3.4

Subtract $\frac{2}{3}$ from $0$: $- \frac{1}{2} \left( - \frac{2}{3} \right)$.

Step:7.4

Simplify the expression.

Step:7.4.1

Multiply $- 1$ by $- 1$: $1 \left( \frac{1}{2} \right) \frac{2}{3}$.

Step:7.4.2

Multiply $\frac{1}{2}$ by $1$: $\frac{1}{2} \cdot \frac{2}{3}$.

Step:7.4.3

Multiply $\frac{1}{2}$ by $\frac{2}{3}$: $\frac{2}{2 \cdot 3}$.

Step:7.4.4

Multiply $2$ by $3$: $\frac{2}{6}$.

Step:7.4.5

Eliminate the common factor of $2$ and $6$.

Step:7.4.5.1

Extract $2$ from $2$: $\frac{2 \cdot 1}{6}$.

Step:7.4.5.2

Cancel the common factors.

Step:7.4.5.2.1

Extract $2$ from $6$: $\frac{2 \cdot 1}{2 \cdot 3}$.

Step:7.4.5.2.2

Cancel the common factor: $\frac{\cancel{2} \cdot 1}{\cancel{2} \cdot 3}$.

Step:7.4.5.2.3

Rephrase the expression: $\frac{1}{3}$.

Step:8

The final result can be presented in different formats.

Exact Form: $\frac{1}{3}$ Decimal Form: $0.333\ldots$

Knowledge Notes:

The problem involves integrating a function using u-substitution, which is a technique for evaluating integrals. The key steps in the u-substitution method include:

  1. Choosing a substitution for $u$ that simplifies the integral.

  2. Differentiating $u$ with respect to $x$ to find $du$.

  3. Changing the limits of integration if the integral is definite.

  4. Rewriting the integral in terms of $u$ and $du$.

  5. Integrating with respect to $u$.

  6. If necessary, converting back to the original variable $x$.

In this problem, the substitution $u = 1 - x^2$ was chosen because the derivative of $-x^2$ is present in the integrand, allowing for a straightforward substitution. The Sum Rule and Power Rule are used for differentiation, and the integral is evaluated using the Power Rule for integration. The limits of integration are also changed to reflect the substitution. After integration, the result is simplified to obtain the final answer.

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