I WILL GIVE YOU A CROWN Ayah purchased 2 meters of ribbon for a project. How many centimeters of ribbon does she have?

Evaluate using integration by parts

PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

$I(n)=\displaystyle\int x^ne^x\,\mathrm dx$

One round of IBP, setting

$u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx$

$\mathrm dv=e^x\,\mathrm dx\implies v=e^x$

gives

$\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx$

$I(n)=x^ne^x-nI(n-1)$

This is called a power-reduction formula. We could try solving for $I(n)$ explicitly, but no need. $n=5$ is small enough to just expand $I(5)$ as much as we need to.

$I(5)=x^5e^x-5I(4)$

$I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)$

$I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)$

$I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)$

$I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))$

Finally,

$I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C$

so we end up with

$I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C$

$I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C$

and the antiderivative is

$\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C$

8 0

2 years ago

Here is a triangular pyramid and its net.

galben [10]

Answer:

a) Area of the base of the pyramid = $15.6\ mm^{2}$

b) Area of one lateral face = $24\ mm^{2}$

c) Lateral Surface Area = $72\ mm^{2}$

d) Total Surface Area = $87.6\ mm^{2}$

Step-by-step explanation:

We are given the following dimensions of the triangular pyramid:

Side of triangular base = 6mm

Height of triangular base = 5.2mm

Base of lateral face (triangular) = 6mm

Height of lateral face (triangular) = 8mm

a) To find Area of base of pyramid:

We know that it is a triangular pyramid and the base is a equilateral triangle. $\text{Area of triangle = } \dfrac{1}{2} \times \text{Base} \times \text{Height} ..... (1)\\$