Math  /  Calculus

QuestionII. Solve the following integrals. Show your solution (3 pts each). 159x2+10x+3dx.01(2x1)6dx\int_{1}^{5} 9 x^{2}+10 x+3 d x . \quad \int_{0}^{1}(2 x-1)^{6} d x

Studdy Solution
Evaluate the definite integral for the second function from 1-1 to 11:
u71411 \left. \frac{u^7}{14} \right|_{-1}^1
Evaluate at the upper limit (1):
(1)714=114 \frac{(1)^7}{14} = \frac{1}{14}
Evaluate at the lower limit (-1):
(1)714=114 \frac{(-1)^7}{14} = \frac{-1}{14}
Subtract the lower limit evaluation from the upper limit evaluation:
114(114)=114+114=214=17 \frac{1}{14} - \left( \frac{-1}{14} \right) = \frac{1}{14} + \frac{1}{14} = \frac{2}{14} = \frac{1}{7}
So, the value of the second integral is:
01(2x1)6dx=17 \int_{0}^{1} (2x - 1)^6 \, dx = \frac{1}{7}
Solution: 15(9x2+10x+3)dx=504 \int_{1}^{5} (9x^2 + 10x + 3) \, dx = 504
01(2x1)6dx=17 \int_{0}^{1} (2x - 1)^6 \, dx = \frac{1}{7}

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