Math Problem StatementSolve the integrals using the Fundamental Theorem of Calculus as specified in the given problems:
1. \(\int_{-2}^{2} \left(x^5 - 4x^3 + 9\right) dx\)
2. \(\int_{0}^{5} \left(3x - 2\right)^2 dx\)
3. \(\int_{-1}^{2} \left(1 + x^2\right)^3 \cdot 2x dx\)
SolutionLet’s solve the integrals step by step as presented in the image. The instructions mention using the Fundamental Theorem of Calculus. Here’s the process for each problem:
Problem 8:
∫ − 2 2 ( x 5 − 4 x 3 + 9 ) d x \int_{-2}^{2} \left(x^5 - 4x^3 + 9\right) dx ∫ − 2 2 ( x 5 − 4 x 3 + 9 ) d x
Solution :
Integrate each term of the polynomial:
∫ x 5 d x = x 6 6 \int x^5 dx = \frac{x^6}{6} ∫ x 5 d x = 6 x 6
∫ − 4 x 3 d x = − 4 x 4 4 = − x 4 \int -4x^3 dx = -\frac{4x^4}{4} = -x^4 ∫ − 4 x 3 d x = − 4 4 x 4 = − x 4
∫ 9 d x = 9 x \int 9 dx = 9x ∫ 9 d x = 9 x
The antiderivative is:
F ( x ) = x 6 6 − x 4 + 9 x F(x) = \frac{x^6}{6} - x^4 + 9x F ( x ) = 6 x 6 − x 4 + 9 x
Apply the Fundamental Theorem of Calculus:
∫ − 2 2 ( x 5 − 4 x 3 + 9 ) d x = F ( 2 ) − F ( − 2 ) \int_{-2}^{2} \left(x^5 - 4x^3 + 9\right) dx = F(2) - F(-2) ∫ − 2 2 ( x 5 − 4 x 3 + 9 ) d x = F ( 2 ) − F ( − 2 )
Calculate F ( 2 ) F(2) F ( 2 ) and F ( − 2 ) F(-2) F ( − 2 ) :
F ( 2 ) = 2 6 6 − 2 4 + 9 ( 2 ) = 64 6 − 16 + 18 = 64 6 + 2 = 74 6 = 37 3 F(2) = \frac{2^6}{6} - 2^4 + 9(2) = \frac{64}{6} - 16 + 18 = \frac{64}{6} + 2 = \frac{74}{6} = \frac{37}{3} F ( 2 ) = 6 2 6 − 2 4 + 9 ( 2 ) = 6 64 − 16 + 18 = 6 64 + 2 = 6 74 = 3 37
F ( − 2 ) = ( − 2 ) 6 6 − ( − 2 ) 4 + 9 ( − 2 ) = 64 6 − 16 − 18 = 64 6 − 34 = 64 6 − 204 6 = − 140 6 = − 70 3 F(-2) = \frac{(-2)^6}{6} - (-2)^4 + 9(-2) = \frac{64}{6} - 16 - 18 = \frac{64}{6} - 34 = \frac{64}{6} - \frac{204}{6} = -\frac{140}{6} = -\frac{70}{3} F ( − 2 ) = 6 ( − 2 ) 6 − ( − 2 ) 4 + 9 ( − 2 ) = 6 64 − 16 − 18 = 6 64 − 34 = 6 64 − 6 204 = − 6 140 = − 3 70
Subtract:
F ( 2 ) − F ( − 2 ) = 37 3 − ( − 70 3 ) = 37 3 + 70 3 = 107 3 F(2) - F(-2) = \frac{37}{3} - \left(-\frac{70}{3}\right) = \frac{37}{3} + \frac{70}{3} = \frac{107}{3} F ( 2 ) − F ( − 2 ) = 3 37 − ( − 3 70 ) = 3 37 + 3 70 = 3 107
Answer :
∫ − 2 2 ( x 5 − 4 x 3 + 9 ) d x = 107 3 \int_{-2}^{2} \left(x^5 - 4x^3 + 9\right) dx = \frac{107}{3} ∫ − 2 2 ( x 5 − 4 x 3 + 9 ) d x = 3 107
Problem 9:
∫ 0 5 ( 3 x − 2 ) 2 d x \int_{0}^{5} \left(3x - 2\right)^2 dx ∫ 0 5 ( 3 x − 2 ) 2 d x
Solution :
Expand the square:
( 3 x − 2 ) 2 = 9 x 2 − 12 x + 4 (3x - 2)^2 = 9x^2 - 12x + 4 ( 3 x − 2 ) 2 = 9 x 2 − 12 x + 4
Integrate each term:
∫ 9 x 2 d x = 3 x 3 \int 9x^2 dx = 3x^3 ∫ 9 x 2 d x = 3 x 3
∫ − 12 x d x = − 6 x 2 \int -12x dx = -6x^2 ∫ − 12 x d x = − 6 x 2
∫ 4 d x = 4 x \int 4 dx = 4x ∫ 4 d x = 4 x
The antiderivative is:
F ( x ) = 3 x 3 − 6 x 2 + 4 x F(x) = 3x^3 - 6x^2 + 4x F ( x ) = 3 x 3 − 6 x 2 + 4 x
Apply the Fundamental Theorem of Calculus:
∫ 0 5 ( 3 x − 2 ) 2 d x = F ( 5 ) − F ( 0 ) \int_{0}^{5} \left(3x - 2\right)^2 dx = F(5) - F(0) ∫ 0 5 ( 3 x − 2 ) 2 d x = F ( 5 ) − F ( 0 )
Calculate F ( 5 ) F(5) F ( 5 ) and F ( 0 ) F(0) F ( 0 ) :
F ( 5 ) = 3 ( 5 ) 3 − 6 ( 5 ) 2 + 4 ( 5 ) = 3 ( 125 ) − 6 ( 25 ) + 20 = 375 − 150 + 20 = 245 F(5) = 3(5)^3 - 6(5)^2 + 4(5) = 3(125) - 6(25) + 20 = 375 - 150 + 20 = 245 F ( 5 ) = 3 ( 5 ) 3 − 6 ( 5 ) 2 + 4 ( 5 ) = 3 ( 125 ) − 6 ( 25 ) + 20 = 375 − 150 + 20 = 245
F ( 0 ) = 3 ( 0 ) 3 − 6 ( 0 ) 2 + 4 ( 0 ) = 0 F(0) = 3(0)^3 - 6(0)^2 + 4(0) = 0 F ( 0 ) = 3 ( 0 ) 3 − 6 ( 0 ) 2 + 4 ( 0 ) = 0
Subtract:
F ( 5 ) − F ( 0 ) = 245 − 0 = 245 F(5) - F(0) = 245 - 0 = 245 F ( 5 ) − F ( 0 ) = 245 − 0 = 245
Answer :
∫ 0 5 ( 3 x − 2 ) 2 d x = 245 \int_{0}^{5} \left(3x - 2\right)^2 dx = 245 ∫ 0 5 ( 3 x − 2 ) 2 d x = 245
Problem 10:
∫ − 1 2 ( 1 + x 2 ) 3 ⋅ 2 x d x \int_{-1}^{2} \left(1 + x^2\right)^3 \cdot 2x \, dx ∫ − 1 2 ( 1 + x 2 ) 3 ⋅ 2 x d x
Solution :
This integral involves substitution:
Let u = 1 + x 2 u = 1 + x^2 u = 1 + x 2 . Then:
d u = 2 x d x du = 2x \, dx d u = 2 x d x
When x = − 1 x = -1 x = − 1 , u = 1 + ( − 1 ) 2 = 2 u = 1 + (-1)^2 = 2 u = 1 + ( − 1 ) 2 = 2
When x = 2 x = 2 x = 2 , u = 1 + 2 2 = 5 u = 1 + 2^2 = 5 u = 1 + 2 2 = 5
Rewrite the integral:
∫ − 1 2 ( 1 + x 2 ) 3 ⋅ 2 x d x = ∫ 2 5 u 3 d u \int_{-1}^{2} \left(1 + x^2\right)^3 \cdot 2x \, dx = \int_{2}^{5} u^3 \, du ∫ − 1 2 ( 1 + x 2 ) 3 ⋅ 2 x d x = ∫ 2 5 u 3 d u
Integrate:
∫ u 3 d u = u 4 4 \int u^3 \, du = \frac{u^4}{4} ∫ u 3 d u = 4 u 4
The antiderivative is:
F ( u ) = u 4 4 F(u) = \frac{u^4}{4} F ( u ) = 4 u 4
Apply the Fundamental Theorem of Calculus:
∫ 2 5 u 3 d u = F ( 5 ) − F ( 2 ) \int_{2}^{5} u^3 \, du = F(5) - F(2) ∫ 2 5 u 3 d u = F ( 5 ) − F ( 2 )
Calculate F ( 5 ) F(5) F ( 5 ) and F ( 2 ) F(2) F ( 2 ) :
F ( 5 ) = 5 4 4 = 625 4 F(5) = \frac{5^4}{4} = \frac{625}{4} F ( 5 ) = 4 5 4 = 4 625
F ( 2 ) = 2 4 4 = 16 4 = 4 F(2) = \frac{2^4}{4} = \frac{16}{4} = 4 F ( 2 ) = 4 2 4 = 4 16 = 4
Subtract:
F ( 5 ) − F ( 2 ) = 625 4 − 4 = 625 4 − 16 4 = 609 4 F(5) - F(2) = \frac{625}{4} - 4 = \frac{625}{4} - \frac{16}{4} = \frac{609}{4} F ( 5 ) − F ( 2 ) = 4 625 − 4 = 4 625 − 4 16 = 4 609
Answer :
∫ − 1 2 ( 1 + x 2 ) 3 ⋅ 2 x d x = 609 4 \int_{-1}^{2} \left(1 + x^2\right)^3 \cdot 2x \, dx = \frac{609}{4} ∫ − 1 2 ( 1 + x 2 ) 3 ⋅ 2 x d x = 4 609
Final Answers:
Problem 8 : 107 3 \frac{107}{3} 3 107
Problem 9 : 245 245 245
Problem 10 : 609 4 \frac{609}{4} 4 609
Would you like detailed explanations for any step? Let me know!
Here are 5 related questions to explore:
How do you perform substitution in definite integrals?
What is the Fundamental Theorem of Calculus and why is it essential?
How do you handle integrals with symmetric limits (e.g., − a -a − a to a a a )?
What is the role of expanding terms in polynomial integration?
How can you confirm an integral solution graphically?
Tip : Always double-check bounds and substitutions when solving definite integrals!