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QuestionMath 110 Course Resources - Definite Integrals Course Packet on the Fundamental Theorem of Calculus
Use the Fundamental Theorem of Calculus to evaluate the following definite integral. ∫277x−28dx=\int_{2}^{7} 7 x-28 d x= □\square Submit Answer 2. [-/1 Points] DETAILS MY NOTES
Math 110 Course Resources - Definite Integrals Course Packet on the Fundamental Theorem of Calculus
Evaluate ∫−33e3x+6x2−3dx\int_{-3}^{3} e^{3 x}+6 x^{2}-3 d x □\square

Studdy Solution

STEP 1

1. We are given two definite integrals to evaluate using the Fundamental Theorem of Calculus.
2. The first integral is ∫27(7x−28) dx \int_{2}^{7} (7x - 28) \, dx .
3. The second integral is ∫−33(e3x+6x2−3) dx \int_{-3}^{3} (e^{3x} + 6x^2 - 3) \, dx .

STEP 2

1. Evaluate the first definite integral using the Fundamental Theorem of Calculus.
2. Evaluate the second definite integral using the Fundamental Theorem of Calculus.

STEP 3

Evaluate the integral ∫(7x−28) dx \int (7x - 28) \, dx :
The antiderivative of 7x 7x is 72x2 \frac{7}{2}x^2 . The antiderivative of −28-28 is −28x-28x.
Thus, the antiderivative of 7x−28 7x - 28 is:
F(x)=72x2−28x F(x) = \frac{7}{2}x^2 - 28x

STEP 4

Apply the Fundamental Theorem of Calculus:
∫27(7x−28) dx=F(7)−F(2) \int_{2}^{7} (7x - 28) \, dx = F(7) - F(2)
Calculate F(7) F(7) :
F(7)=72(7)2−28(7) F(7) = \frac{7}{2}(7)^2 - 28(7) F(7)=72(49)−196 F(7) = \frac{7}{2}(49) - 196 F(7)=3432−196 F(7) = \frac{343}{2} - 196
Calculate F(2) F(2) :
F(2)=72(2)2−28(2) F(2) = \frac{7}{2}(2)^2 - 28(2) F(2)=72(4)−56 F(2) = \frac{7}{2}(4) - 56 F(2)=14−56 F(2) = 14 - 56

STEP 5

Compute the definite integral:
F(7)−F(2)=(3432−196)−(14−56) F(7) - F(2) = \left(\frac{343}{2} - 196\right) - (14 - 56)
F(7)−F(2)=(3432−196)+42 F(7) - F(2) = \left(\frac{343}{2} - 196\right) + 42
Convert 196 196 to a fraction with denominator 2:
F(7)−F(2)=(3432−3922)+42 F(7) - F(2) = \left(\frac{343}{2} - \frac{392}{2}\right) + 42
F(7)−F(2)=−492+42 F(7) - F(2) = \frac{-49}{2} + 42
Convert 42 to a fraction with denominator 2:
F(7)−F(2)=−492+842 F(7) - F(2) = \frac{-49}{2} + \frac{84}{2}
F(7)−F(2)=352 F(7) - F(2) = \frac{35}{2}
The value of the first integral is:
352 \boxed{\frac{35}{2}}

STEP 6

Evaluate the integral ∫(e3x+6x2−3) dx \int (e^{3x} + 6x^2 - 3) \, dx :
The antiderivative of e3x e^{3x} is 13e3x \frac{1}{3}e^{3x} . The antiderivative of 6x2 6x^2 is 2x3 2x^3 . The antiderivative of −3-3 is −3x-3x.
Thus, the antiderivative of e3x+6x2−3 e^{3x} + 6x^2 - 3 is:
G(x)=13e3x+2x3−3x G(x) = \frac{1}{3}e^{3x} + 2x^3 - 3x

STEP 7

Apply the Fundamental Theorem of Calculus:
∫−33(e3x+6x2−3) dx=G(3)−G(−3) \int_{-3}^{3} (e^{3x} + 6x^2 - 3) \, dx = G(3) - G(-3)
Calculate G(3) G(3) :
G(3)=13e9+2(3)3−3(3) G(3) = \frac{1}{3}e^{9} + 2(3)^3 - 3(3) G(3)=13e9+54−9 G(3) = \frac{1}{3}e^{9} + 54 - 9 G(3)=13e9+45 G(3) = \frac{1}{3}e^{9} + 45
Calculate G(−3) G(-3) :
G(−3)=13e−9+2(−3)3−3(−3) G(-3) = \frac{1}{3}e^{-9} + 2(-3)^3 - 3(-3) G(−3)=13e−9−54+9 G(-3) = \frac{1}{3}e^{-9} - 54 + 9 G(−3)=13e−9−45 G(-3) = \frac{1}{3}e^{-9} - 45

STEP 8

Compute the definite integral:
G(3)−G(−3)=(13e9+45)−(13e−9−45) G(3) - G(-3) = \left(\frac{1}{3}e^{9} + 45\right) - \left(\frac{1}{3}e^{-9} - 45\right)
G(3)−G(−3)=13e9+45−13e−9+45 G(3) - G(-3) = \frac{1}{3}e^{9} + 45 - \frac{1}{3}e^{-9} + 45
G(3)−G(−3)=13e9−13e−9+90 G(3) - G(-3) = \frac{1}{3}e^{9} - \frac{1}{3}e^{-9} + 90
The value of the second integral is:
13(e9−e−9)+90 \boxed{\frac{1}{3}(e^{9} - e^{-9}) + 90}

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