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4 years ago

14

The following three shapes are based only on squares, semicircles, and quarter circles. Find the perimeter and the area of each

shaded part. Give your answer as a completely simplified exact value in terms of π (no approximations).

1 answer:

MakcuM [25]4 years ago

6 0

The perimeter = 4 * length of a semicircle of radius 5 == 4* pi * 5 = 20 pi cm

The area of one of the clear parts = 1/2 area of square - area semicircle

= 1/2 * 10^2 - 1/2 pi *5^2 = 50 - 12.5pi

So area of shaded parts = area of square - 4 (50 - 12.5pi)
= 100 - 200 + 100pi = (100pi - 100) cm^2

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If the integral of the product of x squared and e raised to the negative 4 times x power, dx equals the product of negative 1 ov

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Answer:

$A + B + E = 32$

Step-by-step explanation:

Given

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Required

Find $A +B + E$

We have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Using integration by parts

$\int {u} \, dv = uv - \int vdu$

Where

$u = x^2$ and $dv = e^{-4x}dx$

Solve for du (differentiate u)

$du = 2x\ dx$

Solve for v (integrate dv)

$v = -\frac{1}{4}e^{-4x}$

So, we have:

$\int {u} \, dv = uv - \int vdu$

$\int\limits {x^2\cdot e^{-4x}} \, dx = x^2 *-\frac{1}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x} 2xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} - \int -\frac{1}{2}e^{-4x} xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

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Solving

$\int xe^{-4x} dx$

Integration by parts

$u = x$ ---- $du = dx$

$dv = e^{-4x}dx$ ---------- $v = -\frac{1}{4}e^{-4x}$

So:

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} + \int e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}$

So, we have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} [ -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}]$

Open bracket

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} -\frac{x}{8}e^{-4x} -\frac{1}{8}e^{-4x}$

Factor out $e^{-4x}$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{x^2}{4} -\frac{x}{8} -\frac{1}{8}]e^{-4x}$

Rewrite as:

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{4}x^2 -\frac{1}{8}x -\frac{1}{8}]e^{-4x}$

Recall that:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{64}Ax^2 -\frac{1}{64} Bx -\frac{1}{64} E]Ce^{-4x}$

By comparison:

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

$-\frac{1}{8} = -\frac{1}{64}E$

Solve A, B and C

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

Divide by $-x^2$

$\frac{1}{4} = \frac{1}{64}A$

Multiply by 64

$64 * \frac{1}{4} = A$

$A =16$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

Divide by $-x$

$\frac{1}{8} = \frac{1}{64}B$

Multiply by 64

$64 * \frac{1}{8} = \frac{1}{64}B*64$

$B = 8$

$-\frac{1}{8} = -\frac{1}{64}E$

Multiply by -64

$-64 * -\frac{1}{8} = -\frac{1}{64}E * -64$

$E = 8$

So:

$A + B + E = 16 +8+8$

$A + B + E = 32$

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