evaluate $int_{-8}^{0} (-sqrt{-x^{2}-8x})dx$

題目

解題步驟

Understand the question
evaluate $int_{-8}^{0} (-sqrt{-x^{2}-8x})dx$
Explanation
Step1: Complete the square inside root

Rewrite the quadratic expression under the square root by completing the square:
$$-x^2 -8x = -(x^2 +8x) = -(x^2+8x+16-16) = 16-(x+4)^2$$
The integral becomes:
$$\int_{-8}^{0} -\sqrt{16-(x+4)^2} dx$$

Step2: Use substitution $u=x+4$

Let $u = x+4$, so $du = dx$. When $x=-8$, $u=-4$; when $x=0$, $u=4$. The integral transforms to:
$$-\int_{-4}^{4} \sqrt{16-u^2} du$$

Step3: Interpret integral as semicircle area

The integral $\int_{-4}^{4} \sqrt{16-u^2} du$ represents the area of a semicircle with radius $4$ (since $y=\sqrt{16-u^2}$ is the upper half of the circle $u^2+y^2=16$). The area of a full circle is $\pi r^2 = 16\pi$, so the semicircle area is $\frac{1}{2} \times 16\pi = 8\pi$.

Step4: Apply the negative sign

Multiply by the negative sign from the integral:
$$-8\pi$$
Final answer
$-8\pi$

答案

Explanation

Step1: Complete the square inside root

Rewrite the quadratic expression under the square root by completing the square:
$$-x^2 -8x = -(x^2 +8x) = -(x^2+8x+16-16) = 16-(x+4)^2$$
The integral becomes:
$$\int_{-8}^{0} -\sqrt{16-(x+4)^2} dx$$

Step2: Use substitution $u=x+4$

Let $u = x+4$, so $du = dx$. When $x=-8$, $u=-4$; when $x=0$, $u=4$. The integral transforms to:
$$-\int_{-4}^{4} \sqrt{16-u^2} du$$

Step3: Interpret integral as semicircle area

The integral $\int_{-4}^{4} \sqrt{16-u^2} du$ represents the area of a semicircle with radius $4$ (since $y=\sqrt{16-u^2}$ is the upper half of the circle $u^2+y^2=16$). The area of a full circle is $\pi r^2 = 16\pi$, so the semicircle area is $\frac{1}{2} \times 16\pi = 8\pi$.

Step4: Apply the negative sign

Multiply by the negative sign from the integral:
$$-8\pi$$

Answer

$-8\pi$

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Question Analysis

Subject mathematics

Sub Subject calculus

Education Level high school

Difficulty unspecified

Question Type calculation

Multi Question No

Question Count 1

Analysis Status completed

Analyzed At 2026-02-09T19:59:03

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evaluate $int_{-8}^{0} (-sqrt{-x^{2}-8x})dx$

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