Math  /  Calculus

Question3. Let G(x)=1x(t22)dtG(x)=\int_{1}^{x}\left(t^{2}-2\right) d t. Calculate G(1),G(1)G(1), G^{\prime}(1), and G(2)G^{\prime}(2), Then find a formula for G(x)G(x).
4. Find F(0),F(0)F(0), F^{\prime}(0), and F(3)F^{\prime}(3), where F(x)=0xt2+tdtF(x)=\int_{0}^{x} \sqrt{t^{2}+t} d t.
5. Find G(1),G(0)G(1), G^{\prime}(0), and G(π/4)G^{\prime}(\pi / 4), where G(x)=1xtantdtG(x)=\int_{1}^{x} \tan t d t.

Studdy Solution

STEP 1

STEP 2

高层步骤: 1
计算 G(1) G(1) :
G(1)=11(t22)dt=0 G(1) = \int_{1}^{1} (t^2 - 2) \, dt = 0
高层步骤完成: TRUE

STEP 3

高层步骤: 2
根据微积分基本定理,G(x)=t22 G'(x) = t^2 - 2 ,因此:
G(1)=122=1 G'(1) = 1^2 - 2 = -1
高层步骤完成: FALSE

STEP 4

高层步骤: 2
计算 G(2) G'(2) :
G(2)=222=2 G'(2) = 2^2 - 2 = 2
高层步骤完成: TRUE

STEP 5

高层步骤: 3
找出 G(x) G(x) 的公式:
G(x)=1x(t22)dt G(x) = \int_{1}^{x} (t^2 - 2) \, dt
计算不定积分:
=[t332t]1x = \left[ \frac{t^3}{3} - 2t \right]_{1}^{x}
=(x332x)(1332×1) = \left( \frac{x^3}{3} - 2x \right) - \left( \frac{1^3}{3} - 2 \times 1 \right)
=x332x(132) = \frac{x^3}{3} - 2x - \left( \frac{1}{3} - 2 \right)
=x332x+53 = \frac{x^3}{3} - 2x + \frac{5}{3}
高层步骤完成: TRUE

STEP 6

高层步骤: 4
计算 F(0) F(0) :
F(0)=00t2+tdt=0 F(0) = \int_{0}^{0} \sqrt{t^2 + t} \, dt = 0
高层步骤完成: TRUE

STEP 7

高层步骤: 5
根据微积分基本定理,F(x)=x2+x F'(x) = \sqrt{x^2 + x} ,因此:
F(0)=02+0=0 F'(0) = \sqrt{0^2 + 0} = 0
高层步骤完成: FALSE

STEP 8

高层步骤: 5
计算 F(3) F'(3) :
F(3)=32+3=12=23 F'(3) = \sqrt{3^2 + 3} = \sqrt{12} = 2\sqrt{3}
高层步骤完成: TRUE

STEP 9

高层步骤: 6
计算 G(1) G(1) :
G(1)=11tantdt=0 G(1) = \int_{1}^{1} \tan t \, dt = 0
高层步骤完成: TRUE

STEP 10

高层步骤: 7
根据微积分基本定理,G(x)=tanx G'(x) = \tan x ,因此:
G(0)=tan0=0 G'(0) = \tan 0 = 0
高层步骤完成: FALSE

STEP 11

高层步骤: 7
计算 G(π/4) G'(\pi/4) :
G(π/4)=tan(π/4)=1 G'(\pi/4) = \tan(\pi/4) = 1
高层步骤完成: TRUE

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