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To solve the definite integral [tex]\( b = \int_1^4 x^2 \, dx \)[/tex], we will proceed step-by-step. Here's a detailed solution:
1. Identify the integrand and the limits of integration.
The integrand is [tex]\( x^2 \)[/tex] and the limits of integration are from [tex]\( 1 \)[/tex] to [tex]\( 4 \)[/tex].
2. Find the antiderivative of the integrand.
The antiderivative (or indefinite integral) of [tex]\( x^2 \)[/tex] is found by using the power rule for integration. The power rule states that:
[tex]\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \][/tex]
for any real number [tex]\( n \neq -1 \)[/tex].
Applying this rule to [tex]\( x^2 \)[/tex]:
[tex]\[ \int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} \][/tex]
3. Evaluate the definite integral using the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus states that if [tex]\( F(x) \)[/tex] is an antiderivative of [tex]\( f(x) \)[/tex], then:
[tex]\[ \int_a^b f(x) \, dx = F(b) - F(a) \][/tex]
Here, our [tex]\( F(x) = \frac{x^3}{3} \)[/tex]. We need to evaluate this at the upper and lower limits of integration, [tex]\( x = 4 \)[/tex] and [tex]\( x = 1 \)[/tex], respectively.
[tex]\[ F(4) = \frac{4^3}{3} = \frac{64}{3} \][/tex]
[tex]\[ F(1) = \frac{1^3}{3} = \frac{1}{3} \][/tex]
4. Subtract the value of the antiderivative at the lower limit from the value at the upper limit.
[tex]\[ \int_1^4 x^2 \, dx = F(4) - F(1) = \frac{64}{3} - \frac{1}{3} \][/tex]
5. Simplify the expression.
Combine the fractions:
[tex]\[ \frac{64}{3} - \frac{1}{3} = \frac{64 - 1}{3} = \frac{63}{3} = 21 \][/tex]
Therefore, the value of the definite integral [tex]\( b = \int_1^4 x^2 \, dx \)[/tex] is:
[tex]\[ b = 21 \][/tex]
1. Identify the integrand and the limits of integration.
The integrand is [tex]\( x^2 \)[/tex] and the limits of integration are from [tex]\( 1 \)[/tex] to [tex]\( 4 \)[/tex].
2. Find the antiderivative of the integrand.
The antiderivative (or indefinite integral) of [tex]\( x^2 \)[/tex] is found by using the power rule for integration. The power rule states that:
[tex]\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \][/tex]
for any real number [tex]\( n \neq -1 \)[/tex].
Applying this rule to [tex]\( x^2 \)[/tex]:
[tex]\[ \int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} \][/tex]
3. Evaluate the definite integral using the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus states that if [tex]\( F(x) \)[/tex] is an antiderivative of [tex]\( f(x) \)[/tex], then:
[tex]\[ \int_a^b f(x) \, dx = F(b) - F(a) \][/tex]
Here, our [tex]\( F(x) = \frac{x^3}{3} \)[/tex]. We need to evaluate this at the upper and lower limits of integration, [tex]\( x = 4 \)[/tex] and [tex]\( x = 1 \)[/tex], respectively.
[tex]\[ F(4) = \frac{4^3}{3} = \frac{64}{3} \][/tex]
[tex]\[ F(1) = \frac{1^3}{3} = \frac{1}{3} \][/tex]
4. Subtract the value of the antiderivative at the lower limit from the value at the upper limit.
[tex]\[ \int_1^4 x^2 \, dx = F(4) - F(1) = \frac{64}{3} - \frac{1}{3} \][/tex]
5. Simplify the expression.
Combine the fractions:
[tex]\[ \frac{64}{3} - \frac{1}{3} = \frac{64 - 1}{3} = \frac{63}{3} = 21 \][/tex]
Therefore, the value of the definite integral [tex]\( b = \int_1^4 x^2 \, dx \)[/tex] is:
[tex]\[ b = 21 \][/tex]
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