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3 years ago

10

How would you integrate (2x-1)(6x+5)? is there a product rule for integration?

1 answer:

NikAS [45]3 years ago

4 0

No, there is not a product rule for integration.

To integrate (2x - 1)(6x + 5), start expanding the product:

(2x - 1)(6x + 5) = 12x^2 + 10x - 6x - 5 = 12x^2 + 4x - 5.

Now, I guess you know how to integrate that:

∫ (12x^2 + 4x - 5) dx = ∫12x^2 dx + ∫4x dx + ∫ (-5) dx = 4x^3 + 2x^2 - 5x + C

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Here are the vertices of rectangle FROG: (-2,5),(-2,1),(6,5),(6,1). Find the perimeter of this rectangle. If you get stuck, try

ryzh [129]

Answer:

Part 1) The perimeter of rectangle is equal to 24 units

Part 2) The area of rectangle is equal to 32 square units

Step-by-step explanation:

Part 1) Find the perimeter of rectangle

we know that

The perimeter of rectangle is equal to

$P=2(L+W)$

where

L is the length of rectangle

W is the width of rectangle

we have

$F(-2,5),R(-2,1),O(6,1),G(6,5)$

Plot the figure to better understand the problem

using a graphing tool

see the attached figure

Remember that in a rectangle opposite sides are congruent and the measure of each interior angle is equal to 90 degrees

so

$FG=RO=L\\RF=OG=W$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance FG

$F(-2,5),G(6,5)$

substitute the values

$d=\sqrt{(5-5)^{2}+(6+2)^{2}}$

$d=\sqrt{(0)^{2}+(8)^{2}}$

$FG=8\ units$

step 2

Find the distance RF

$R(-2,1),F(-2,5)$

substitute the values

$d=\sqrt{(5-1)^{2}+(-2+2)^{2}}$

$d=\sqrt{(4)^{2}+(0)^{2}}$

$RF=4\ units$

step 3

Find the perimeter

$P=2(L+W)$

we have

$FG=RO=L=8\ units\\RF=OG=W=4\ units$

substitute

$P=2(8+4)=24\ units$

Part 2) Find the area of rectangle FROG

we know that

The area of rectangle is equal to

$A=LW$

we have

$FG=RO=L=8\ units\\RF=OG=W=4\ units$

substitute

$A=(8)(4)=32\ units^2$

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t=4

explanation

2(t+1)=10

2t+2=10

2t=8

t=4

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