Math / Calculus

QuestionEvaluate the integral: $\int_{0}^{0.4} \frac{d x}{\sqrt{1-x^{2}}}$

Studdy Solution

STEP 1

1. The integral $\int \frac{dx}{\sqrt{1-x^2}}$ is a standard integral that can be solved using trigonometric substitution or recognizing it as a derivative of an inverse trigonometric function.
2. The limits of integration are from 0 to 0.4.

STEP 2

1. Recognize the integral as the derivative of an inverse trigonometric function.
2. Evaluate the antiderivative at the upper and lower limits.
3. Compute the definite integral.

STEP 3

Recognize that the integral $\int \frac{dx}{\sqrt{1-x^2}}$ is the derivative of the inverse sine function, $\sin^{-1}(x)$ . Therefore, the antiderivative is:
$\sin^{-1}(x) + C$

STEP 4

Evaluate the antiderivative at the upper limit $x = 0.4$ and the lower limit $x = 0$ :
$\left[ \sin^{-1}(x) \right]_{0}^{0.4} = \sin^{-1}(0.4) - \sin^{-1}(0)$

STEP 5

Compute the values of the inverse sine function:
1. $\sin^{-1}(0.4)$ is the angle whose sine is 0.4.
2. $\sin^{-1}(0)$ is the angle whose sine is 0, which is 0.
Thus, the definite integral is:
$\sin^{-1}(0.4) - 0 = \sin^{-1}(0.4)$
The value of the integral is:
$\boxed{\sin^{-1}(0.4)}$

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QuestionEvaluate the integral: $\int_{0}^{0.4} \frac{d x}{\sqrt{1-x^{2}}}$

Studdy Solution

STEP 1

1. The integral $\int \frac{dx}{\sqrt{1-x^2}}$ is a standard integral that can be solved using trigonometric substitution or recognizing it as a derivative of an inverse trigonometric function.
2. The limits of integration are from 0 to 0.4.

STEP 2

1. Recognize the integral as the derivative of an inverse trigonometric function.
2. Evaluate the antiderivative at the upper and lower limits.
3. Compute the definite integral.

STEP 3

Recognize that the integral $\int \frac{dx}{\sqrt{1-x^2}}$ is the derivative of the inverse sine function, $\sin^{-1}(x)$ . Therefore, the antiderivative is:
$\sin^{-1}(x) + C$

STEP 4

Evaluate the antiderivative at the upper limit $x = 0.4$ and the lower limit $x = 0$ :
$\left[ \sin^{-1}(x) \right]_{0}^{0.4} = \sin^{-1}(0.4) - \sin^{-1}(0)$

STEP 5

Compute the values of the inverse sine function:
1. $\sin^{-1}(0.4)$ is the angle whose sine is 0.4.
2. $\sin^{-1}(0)$ is the angle whose sine is 0, which is 0.
Thus, the definite integral is:
$\sin^{-1}(0.4) - 0 = \sin^{-1}(0.4)$
The value of the integral is:
$\boxed{\sin^{-1}(0.4)}$

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QuestionEvaluate the integral: ∫00.4dx1−x2\int_{0}^{0.4} \frac{d x}{\sqrt{1-x^{2}}}∫00.4​1−x2​dx​

Studdy Solution

STEP 1

1. The integral ∫dx1−x2\int \frac{dx}{\sqrt{1-x^2}}∫1−x2​dx​ is a standard integral that can be solved using trigonometric substitution or recognizing it as a derivative of an inverse trigonometric function.2. The limits of integration are from 0 to 0.4.

STEP 2

1. Recognize the integral as the derivative of an inverse trigonometric function.2. Evaluate the antiderivative at the upper and lower limits.3. Compute the definite integral.

STEP 3

Recognize that the integral ∫dx1−x2\int \frac{dx}{\sqrt{1-x^2}}∫1−x2​dx​ is the derivative of the inverse sine function, sin⁡−1(x)\sin^{-1}(x)sin−1(x). Therefore, the antiderivative is:sin⁡−1(x)+C \sin^{-1}(x) + C sin−1(x)+C

STEP 4

Evaluate the antiderivative at the upper limit x=0.4x = 0.4x=0.4 and the lower limit x=0x = 0x=0:[sin⁡−1(x)]00.4=sin⁡−1(0.4)−sin⁡−1(0) \left[ \sin^{-1}(x) \right]_{0}^{0.4} = \sin^{-1}(0.4) - \sin^{-1}(0) [sin−1(x)]00.4​=sin−1(0.4)−sin−1(0)

STEP 5

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QuestionEvaluate the integral: $\int_{0}^{0.4} \frac{d x}{\sqrt{1-x^{2}}}$

1. The integral $\int \frac{dx}{\sqrt{1-x^2}}$ is a standard integral that can be solved using trigonometric substitution or recognizing it as a derivative of an inverse trigonometric function.
2. The limits of integration are from 0 to 0.4.

1. Recognize the integral as the derivative of an inverse trigonometric function.
2. Evaluate the antiderivative at the upper and lower limits.
3. Compute the definite integral.

Recognize that the integral $\int \frac{dx}{\sqrt{1-x^2}}$ is the derivative of the inverse sine function, $\sin^{-1}(x)$ . Therefore, the antiderivative is:
$\sin^{-1}(x) + C$

Evaluate the antiderivative at the upper limit $x = 0.4$ and the lower limit $x = 0$ :
$\left[ \sin^{-1}(x) \right]_{0}^{0.4} = \sin^{-1}(0.4) - \sin^{-1}(0)$