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Math  /  Calculus

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6. [-/1 Points] DETAILS MY NOTES 25%25 \% ASK YOUR TEACHER
Evaluate the integral. ∫01x(6x3+7x4)dx\int_{0}^{1} x(6 \sqrt[3]{x}+7 \sqrt[4]{x}) d x □\square Submit Answer
7. [-/1 Points] DETAILS MY NOTES ASK YOUR TEACHER
Evaluate the integral. ∫−2−1(6y5+4y5)dy\int_{-2}^{-1}\left(6 y^{5}+\frac{4}{y^{5}}\right) d y □\square Submit Answer
8. [-/1 Points] DETAILS MY NOTES ASK YOUR TEACHER
Evaluate the integral. ∫01(x12+12x)dx\int_{0}^{1}\left(x^{12}+12^{x}\right) d x □\square Submit Answer
9. [-/1 Points] DETAILS MY NOTES ASK YOUR TEACHER
Evaluate the integral. ∫0π/45+7cos⁡2θcos⁡2θdθ\int_{0}^{\pi / 4} \frac{5+7 \cos ^{2} \theta}{\cos ^{2} \theta} d \theta □\square Submit Answer
10. [-/1 Points] DETAILS MY NOTES ASK YOUR TEACHER
Evaluate the integral, ∫12y+6y8y3dy\int_{1}^{2} \frac{y+6 y^{8}}{y^{3}} d y □\square Submit Answer

Studdy Solution

STEP 1

1. We are dealing with a definite integral from 0 to 1.
2. The integrand can be simplified by rewriting the roots as fractional exponents.

STEP 2

1. Simplify the integrand.
2. Integrate term by term.
3. Evaluate the definite integral.
4. 1. Simplify the integrand.
2. Integrate term by term.
3. Evaluate the definite integral.
STEP_1: High_Level_Step: 1 Rewrite the integrand:
6y5+4y5=6y5+4y−5 6y^5 + \frac{4}{y^5} = 6y^5 + 4y^{-5}
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2 Integrate each term separately:
∫6y5 dy=6⋅y66=y6 \int 6y^5 \, dy = 6 \cdot \frac{y^{6}}{6} = y^6
∫4y−5 dy=4⋅y−4−4=−y−4 \int 4y^{-5} \, dy = 4 \cdot \frac{y^{-4}}{-4} = -y^{-4}
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 3 Evaluate the definite integral from -2 to -1:
[y6−y−4]−2−1 \left[ y^6 - y^{-4} \right]_{-2}^{-1}
=((−1)6−(−1)−4)−((−2)6−(−2)−4) = \left( (-1)^6 - (-1)^{-4} \right) - \left( (-2)^6 - (-2)^{-4} \right)
=(1−1)−(64−116) = \left( 1 - 1 \right) - \left( 64 - \frac{1}{16} \right)
=0−(64−116) = 0 - \left( 64 - \frac{1}{16} \right)
=−64+116 = -64 + \frac{1}{16}
=−102416+116 = -\frac{1024}{16} + \frac{1}{16}
=−102316 = -\frac{1023}{16}
High_Level_Step_Completed: TRUE
The value of the integral is:
−102316 \boxed{-\frac{1023}{16}}
Let's move on to the next problem.
### Problem 8: Evaluate the integral ∫01(x12+12x) dx\int_{0}^{1}\left(x^{12}+12^{x}\right) \, dx
_ASSUMPTIONS_:
1. We are dealing with a definite integral from 0 to 1.
2. The integrand consists of a polynomial and an exponential function.
_HIGH_LEVEL_APPROACH_:
1. Integrate the polynomial term.
2. Integrate the exponential term.
3. Evaluate the definite integral.
STEP_1: High_Level_Step: 1 Integrate the polynomial term:
∫x12 dx=x1313 \int x^{12} \, dx = \frac{x^{13}}{13}
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2 Integrate the exponential term:
The integral of 12x 12^x is 12xln⁡(12) \frac{12^x}{\ln(12)} .
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 3 Evaluate the definite integral from 0 to 1:
[x1313+12xln⁡(12)]01 \left[ \frac{x^{13}}{13} + \frac{12^x}{\ln(12)} \right]_{0}^{1}
=(11313+121ln⁡(12))−(01313+120ln⁡(12)) = \left( \frac{1^{13}}{13} + \frac{12^1}{\ln(12)} \right) - \left( \frac{0^{13}}{13} + \frac{12^0}{\ln(12)} \right)
=(113+12ln⁡(12))−(0+1ln⁡(12)) = \left( \frac{1}{13} + \frac{12}{\ln(12)} \right) - \left( 0 + \frac{1}{\ln(12)} \right)
=113+12ln⁡(12)−1ln⁡(12) = \frac{1}{13} + \frac{12}{\ln(12)} - \frac{1}{\ln(12)}
=113+11ln⁡(12) = \frac{1}{13} + \frac{11}{\ln(12)}
High_Level_Step_Completed: TRUE
The value of the integral is:
113+11ln⁡(12) \boxed{\frac{1}{13} + \frac{11}{\ln(12)}}
Let's move on to the next problem.
### Problem 9: Evaluate the integral ∫0π/45+7cos⁡2θcos⁡2θ dθ\int_{0}^{\pi / 4} \frac{5+7 \cos ^{2} \theta}{\cos ^{2} \theta} \, d \theta
_ASSUMPTIONS_:
1. We are dealing with a definite integral from 0 to π/4\pi/4.
2. The integrand can be simplified using trigonometric identities.
_HIGH_LEVEL_APPROACH_:
1. Simplify the integrand.
2. Integrate the simplified expression.
3. Evaluate the definite integral.
STEP_1: High_Level_Step: 1 Simplify the integrand:
5+7cos⁡2θcos⁡2θ=5cos⁡2θ+7 \frac{5+7 \cos ^{2} \theta}{\cos ^{2} \theta} = \frac{5}{\cos^2 \theta} + 7
=5sec⁡2θ+7 = 5 \sec^2 \theta + 7
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2 Integrate each term separately:
∫5sec⁡2θ dθ=5tan⁡θ \int 5 \sec^2 \theta \, d\theta = 5 \tan \theta
∫7 dθ=7θ \int 7 \, d\theta = 7\theta
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 3 Evaluate the definite integral from 0 to π/4\pi/4:
[5tan⁡θ+7θ]0π/4 \left[ 5 \tan \theta + 7\theta \right]_{0}^{\pi/4}
=(5tan⁡π4+7⋅π4)−(5tan⁡0+7⋅0) = \left( 5 \tan \frac{\pi}{4} + 7 \cdot \frac{\pi}{4} \right) - \left( 5 \tan 0 + 7 \cdot 0 \right)
=(5⋅1+7π4)−0 = \left( 5 \cdot 1 + \frac{7\pi}{4} \right) - 0
=5+7π4 = 5 + \frac{7\pi}{4}
High_Level_Step_Completed: TRUE
The value of the integral is:
5+7π4 \boxed{5 + \frac{7\pi}{4}}
Let's move on to the final problem.
### Problem 10: Evaluate the integral ∫12y+6y8y3 dy\int_{1}^{2} \frac{y+6 y^{8}}{y^{3}} \, dy
_ASSUMPTIONS_:
1. We are dealing with a definite integral from 1 to 2.
2. The integrand can be simplified by dividing each term by y3 y^3 .
_HIGH_LEVEL_APPROACH_:
1. Simplify the integrand.
2. Integrate term by term.
3. Evaluate the definite integral.
STEP_1: High_Level_Step: 1 Simplify the integrand:
y+6y8y3=y−2+6y5 \frac{y+6 y^{8}}{y^{3}} = y^{-2} + 6y^{5}
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2 Integrate each term separately:
∫y−2 dy=y−1−1=−y−1 \int y^{-2} \, dy = \frac{y^{-1}}{-1} = -y^{-1}
∫6y5 dy=6⋅y66=y6 \int 6y^{5} \, dy = 6 \cdot \frac{y^{6}}{6} = y^6
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 3 Evaluate the definite integral from 1 to 2:
[−y−1+y6]12 \left[ -y^{-1} + y^6 \right]_{1}^{2}
=(−12+26)−(−11+16) = \left( -\frac{1}{2} + 2^6 \right) - \left( -\frac{1}{1} + 1^6 \right)
=(−12+64)−(−1+1) = \left( -\frac{1}{2} + 64 \right) - \left( -1 + 1 \right)
=(−12+64)+0 = \left( -\frac{1}{2} + 64 \right) + 0
=−12+1282 = \frac{-1}{2} + \frac{128}{2}
=1272 = \frac{127}{2}
High_Level_Step_Completed: TRUE
The value of the integral is:
1272 \boxed{\frac{127}{2}}

STEP 3

Rewrite the roots as fractional exponents:
6x3=6x1/3 6 \sqrt[3]{x} = 6x^{1/3} 7x4=7x1/4 7 \sqrt[4]{x} = 7x^{1/4}
Thus, the integrand becomes:
x(6x1/3+7x1/4)=6x1+1/3+7x1+1/4=6x4/3+7x5/4 x(6x^{1/3} + 7x^{1/4}) = 6x^{1 + 1/3} + 7x^{1 + 1/4} = 6x^{4/3} + 7x^{5/4}

STEP 4

Integrate each term separately:
∫6x4/3 dx=6⋅x4/3+14/3+1=6⋅x7/37/3=187x7/3 \int 6x^{4/3} \, dx = 6 \cdot \frac{x^{4/3 + 1}}{4/3 + 1} = 6 \cdot \frac{x^{7/3}}{7/3} = \frac{18}{7} x^{7/3}
∫7x5/4 dx=7⋅x5/4+15/4+1=7⋅x9/49/4=289x9/4 \int 7x^{5/4} \, dx = 7 \cdot \frac{x^{5/4 + 1}}{5/4 + 1} = 7 \cdot \frac{x^{9/4}}{9/4} = \frac{28}{9} x^{9/4}

STEP 5

Evaluate the definite integral from 0 to 1:
[187x7/3+289x9/4]01 \left[ \frac{18}{7} x^{7/3} + \frac{28}{9} x^{9/4} \right]_{0}^{1}
=(187⋅17/3+289⋅19/4)−(187⋅07/3+289⋅09/4) = \left( \frac{18}{7} \cdot 1^{7/3} + \frac{28}{9} \cdot 1^{9/4} \right) - \left( \frac{18}{7} \cdot 0^{7/3} + \frac{28}{9} \cdot 0^{9/4} \right)
=187+289 = \frac{18}{7} + \frac{28}{9}
=16263+19663 = \frac{162}{63} + \frac{196}{63}
=35863 = \frac{358}{63}
The value of the integral is:
35863 \boxed{\frac{358}{63}}
Let's move on to the next problem.
### Problem 7: Evaluate the integral ∫−2−1(6y5+4y5) dy\int_{-2}^{-1}\left(6 y^{5}+\frac{4}{y^{5}}\right) \, dy
_ASSUMPTIONS_:
1. We are dealing with a definite integral from -2 to -1.
2. The integrand is a sum of a polynomial and a rational function.

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