elimination y+z=-4, 5x+4y-16z=-6, x-y+5z=6

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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∫₁²√x³−1dx, n=3

Solution

1.23762…

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Solve by:

Solve by Simpson's rule

Solve by midpoint Riemann sum

Solve by left point Riemann sum

Solve by right point Riemann sum

Solve by trapezoidal rule

One step at a time

Simpson's Rule (Approximation)

∫_a^bf(x)dx ≈ Δ x3 (f(x₀)+4f(x₁)+2f(x₂)+4f(x₃)+…+4f(x_n−2)+f(x_n−1)),
where Δ x = b−an , n is even

Apply Simpson's Formula

=19 (f(x₀)+4f(x₁)+2f(x₂)+f(x₃))

Calculate Sub Intervals

=19 (0+4√1119 +14√69 +√7)

=1.23762…

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Solve by: midpoint Riemann sum

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

elimination y+z=-4, 5x+4y-16z=-6, x-y+5z=6

Solution

Solution steps

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