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lora16 [44]
3 years ago
9

Use the value of the first integral I to evaluate the two given integrals. I = integral^3_0 (x^3 - 4x)dx = 9/4 A. integral^3_0 (

8x - 2x^3)dx B. integral^0_3 (4x - x^3)dx C. integral^3_0 (8x - 2x^3)dx =
Mathematics
1 answer:
Troyanec [42]3 years ago
4 0

Answer:

A) -9/2

B) 9/4

C) -9/2, same as A)

Step-by-step explanation:

We are given that I=\int_0^3 x^3-4x dx=9/4. We use the properties of integrals to write the new integrals in terms of I.

A) \int_0^3 8x-2x^3 dx=\int_0^3 -2(x^3-4x) dx=-2\int_0^3 x^3-4x dx=-2I=-9/2. We have used that ∫cf dx=c∫f dx.

B) \int_3^0 4x - x^3 dx=-\int_0^3 (4x-x^3) dx=\int_0^3-(4x-x^3) dx=\int_0^3 x^3-4x dx=I=9/4. Here we used that reversing the limits of integration changes the sign of the integral.

C) It's the same integral in A)

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Remove parentheses
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The base of an isosceles triangle is one and a half times the length of the other two sides.A smaller triangle has a perimeter t
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                                                          = 2x + x + \frac{x}{2}

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Since, smaller triangle has a perimeter that is half the perimeter of the larger triangle,

Length of each corresponding side of smaller triangle will be half of the larger triangle.

Length of sides of the smaller triangle = \frac{x}{2},\frac{x}{2},(1\frac{1}{2})\frac{x}{2}

Perimeter of the smaller triangle = \frac{x}{2}+\frac{x}{2}+(1\frac{1}{2})\frac{x}{2}

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                                                      = \frac{3}{2}x+\frac{x}{4}

                                                      = \frac{6}{4}x+\frac{x}{4}

                                                      = \frac{7}{4}x

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Therefore, expression that represents the perimeter of the smaller triangle is (1.75x)        

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