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Math  /  Calculus

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

Studdy Solution

STEP 1

1. The first integral is a definite integral of a polynomial function from 1 to 5.
2. The second integral is a definite integral of a polynomial function raised to a power from 0 to 1.
3. Standard techniques of integration, such as power rule and substitution, will be used.
4. For definite integrals, the Fundamental Theorem of Calculus will be applied.

STEP 2

1. Integrate the polynomial function for the first integral.
2. Evaluate the definite integral for the first function.
3. Simplify the integrand of the second integral using binomial expansion or substitution.
4. Integrate the simplified form of the second function.
5. Evaluate the definite integral for the second function.

STEP 3

Integrate the polynomial function for the first integral.
∫(9x2+10x+3) dx \int (9x^2 + 10x + 3) \, dx

STEP 4

Apply the power rule to each term in the polynomial:
∫9x2 dx+∫10x dx+∫3 dx \int 9x^2 \, dx + \int 10x \, dx + \int 3 \, dx
=9∫x2 dx+10∫x dx+3∫1 dx = 9 \int x^2 \, dx + 10 \int x \, dx + 3 \int 1 \, dx

STEP 5

Integrate each term:
9∫x2 dx=9⋅x33=3x3 9 \int x^2 \, dx = 9 \cdot \frac{x^3}{3} = 3x^3
10∫x dx=10⋅x22=5x2 10 \int x \, dx = 10 \cdot \frac{x^2}{2} = 5x^2
3∫1 dx=3x 3 \int 1 \, dx = 3x
Therefore, the antiderivative is:
3x3+5x2+3x 3x^3 + 5x^2 + 3x

STEP 6

Evaluate the definite integral for the first function from 1 to 5:
∫15(3x3+5x2+3x) dx \int_{1}^{5} (3x^3 + 5x^2 + 3x) \, dx

STEP 7

Apply the Fundamental Theorem of Calculus:
[3x3+5x2+3x]15 \left[ 3x^3 + 5x^2 + 3x \right]_1^5
First, evaluate at the upper limit (5):
3(5)3+5(5)2+3(5) 3(5)^3 + 5(5)^2 + 3(5)
=3(125)+5(25)+15 = 3(125) + 5(25) + 15
=375+125+15=515 = 375 + 125 + 15 = 515
Then, evaluate at the lower limit (1):
3(1)3+5(1)2+3(1) 3(1)^3 + 5(1)^2 + 3(1)
=3(1)+5(1)+3(1) = 3(1) + 5(1) + 3(1)
=3+5+3=11 = 3 + 5 + 3 = 11
Subtract the lower limit evaluation from the upper limit evaluation:
515−11=504 515 - 11 = 504
So, the value of the first integral is:
∫15(9x2+10x+3) dx=504 \int_{1}^{5} (9x^2 + 10x + 3) \, dx = 504

STEP 8

Simplify the integrand of the second integral using binomial expansion or substitution.
∫01(2x−1)6 dx \int_{0}^{1} (2x - 1)^6 \, dx
Consider substitution: Let u=2x−1u = 2x - 1. Then, du=2dxdu = 2dx or dx=du2dx = \frac{du}{2}.

STEP 9

Change the limits of integration accordingly: when x=0x = 0, u=−1u = -1; and when x=1x = 1, u=1u = 1.
∫−11u6⋅12 du \int_{-1}^{1} u^6 \cdot \frac{1}{2} \, du
=12∫−11u6 du = \frac{1}{2} \int_{-1}^{1} u^6 \, du

STEP 10

Integrate the simplified form of the second function:
12∫u6 du \frac{1}{2} \int u^6 \, du
Apply the power rule:
12⋅u77=u714 \frac{1}{2} \cdot \frac{u^7}{7} = \frac{u^7}{14}

STEP 11

Evaluate the definite integral for the second function from −1-1 to 11:
u714∣−11 \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(2x−1)6 dx=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(2x−1)6 dx=17 \int_{0}^{1} (2x - 1)^6 \, dx = \frac{1}{7}

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