Calculate the definite integral ∫ (1 → 4) LNX / Radix x DX

Calculate the definite integral ∫ (1 → 4) LNX / Radix x DX

Let x = t,
When x ∈ [1,4], TX ∈ [1,2]
∫ (1 → 4) LNX / Radix x DX = ∫ (1 → 2) 2lnt / T * 2tdt = 4 ∫ (1 → 2) lntdt = 4T * LNT | (1 → 2) - 4 ∫ (1 → 2) tdlnt = 4 * 2ln2-4 * 1 * ln1-4 ∫ (1 → 2) t * 1 / TDT = 8ln2-4 ∫ (1 → 2) DT = 8ln2-4t | (1 → 2)
=8ln2-4*2+4*1=8ln2-4

Calculate the definite integral ∫ (1 → radical 3) [1 / {x}] DX

This problem is replaced by trigonometry
Let x = tant, then DX = sec? TDT
∵x∈[1,√3]
ν let t ∈ [π / 4, π / 3] (in this interval, x increases with T, sect ≥ 0)
Original integral = ∫ (π / 4, π / 3) sec? TDT / (tan? T · sect)
=∫(π/4,π/3) sectdt/tan²t
=∫(π/4,π/3) dt/(tan²t·cost)
=∫(π/4,π/3) costdt/sin²t
=∫(π/4,π/3) d(sint)/sin²t
=[-1/sint]|(π/4,π/3)
=√2-2/√3
=√2-√6/3
I hope my answer will help you

Calculate the definite integral ∫ (root 2 above, 1 below) x / Radix 4-x ^ 2 * DX

∫[1,√2]x/√（4-x^2 ）dx
=-1/2∫[1,√2]1/√（4-x^2 ）d(4-x^2)
=-√（4-x^2 ）[1,√2]
=√3－√2

Using the geometric meaning of definite integral to find ∫ upper 6, lower 0, root sign 9 - (x-3) ^ 2DX

y = √[9 - (x - 3)²]
(x - 3)² + y² = 3²
Center (3,0), radius 3
From 0 to 6, just around a semicircle
So ∫ (0 → 6) √ [9 - (x - 3) 2] DX = 1 / 2 · π (3) 2 = 9 π / 2

Definite integral upper limit 1 lower limit 1 root sign 1-x square

Let x = Sint, when x = - 1, t = - π / 2, x = 1, t = π / 2, when x = 1, t = π / 2, x = 1, t = π / 2, t = π / 2, so the original formula = ∫ (- π / 2, π / 2) costdsint = (- (π / 2, π / 2) cos s (- π / 2, π / 2) cos t = 1 / 2 (- (π / 2, π / 2) (1 + cos 2t) DT = 1 / 2 (T + 1 / 2 sin2t), t = π / 2, t = π / 2, t, t = π / 2, the original formula = (- (π / 2 it's a good idea

According to the geometric meaning of definite integral, ∫ One Zero 4−x2dx=______ .

From the geometric meaning of definite integral, we can know that:
A kind of
One
Zero
4 − x2dx is the area of the curved trapezoid oabc in the shaded portion as shown in the figure,
Where B (1,
3），∠BOC=30°
Therefore
One
Zero
4 − x2dx = s sector BOC + s △ AOB = π
3+
Three
Two
So the answer is: π
3+
Three
Two

Calculate the definite integral 1,0 (1 / radical 1 + x) DX 1,0 is the number above and below the definite integral

Definite integral 1,0 (1 / radical 1 + x) DX
Let t = 1 + X, then 1

How to use the geometric meaning of definite integral to find ((x) under {radical sign The definite integral (b > A) over a to B under the radical ((x-a) (b-X))} has to be solved in geometric sense. Sorry, I can't edit the formula with word. I hope it's better to bring the formula

See figure

Definite integral of 1-x ^ 2 from 0 to 1 under x + 1 / radical

 

Find the definite integral 1 / x ^ 2 (1 + x ^ 2) ^ 1 / 2, upper limit, root sign, 3, lower limit, 1

Let x = Tan θ, DX = sec 2 θ D θ, X ∈ [1, √ 3] → θ∈ [π / 4, π / 3] ∫ (1 ~ √ (1 + X?) DX = ∫ (π / 4 ~ π / 3) sec θ / (Tan θ sec θ) d θ = ∫ (π / 4 ~ π / 3) 1 / cos θ· cos θ / sin