Math / Calculus

QuestionEvaluate exactly, using the Fundamental Theorem of Calculus: $\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x=$ $\square$

Studdy Solution

STEP 1

1. We are given the definite integral $\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x$ .
2. We will use the Fundamental Theorem of Calculus to evaluate the integral.

STEP 2

1. Identify the integrand and split the integral if necessary.
2. Find the antiderivative of the integrand.
3. Apply the Fundamental Theorem of Calculus to evaluate the definite integral.
4. Simplify the result.

STEP 3

Identify the integrand:
The integrand is $\frac{x^{2}}{9} + 6x$ .

STEP 4

Find the antiderivative of the integrand:
The antiderivative of $\frac{x^{2}}{9}$ is $\frac{x^3}{27}$ .
The antiderivative of $6x$ is $3x^2$ .
Therefore, the antiderivative of the entire integrand is:
$F(x) = \frac{x^3}{27} + 3x^2$

STEP 5

Apply the Fundamental Theorem of Calculus:
$\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x = F(a) - F(0)$
Substitute the antiderivative into the expression:
$F(a) = \frac{a^3}{27} + 3a^2$
$F(0) = \frac{0^3}{27} + 3(0)^2 = 0$

STEP 6

Evaluate the expression:
$F(a) - F(0) = \left(\frac{a^3}{27} + 3a^2\right) - 0$
Simplify:
$\frac{a^3}{27} + 3a^2$
The exact value of the integral is:
$\boxed{\frac{a^3}{27} + 3a^2}$

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QuestionEvaluate exactly, using the Fundamental Theorem of Calculus: $\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x=$ $\square$

Studdy Solution

STEP 1

1. We are given the definite integral $\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x$ .
2. We will use the Fundamental Theorem of Calculus to evaluate the integral.

STEP 2

1. Identify the integrand and split the integral if necessary.
2. Find the antiderivative of the integrand.
3. Apply the Fundamental Theorem of Calculus to evaluate the definite integral.
4. Simplify the result.

STEP 3

Identify the integrand:
The integrand is $\frac{x^{2}}{9} + 6x$ .

STEP 4

Find the antiderivative of the integrand:
The antiderivative of $\frac{x^{2}}{9}$ is $\frac{x^3}{27}$ .
The antiderivative of $6x$ is $3x^2$ .
Therefore, the antiderivative of the entire integrand is:
$F(x) = \frac{x^3}{27} + 3x^2$

STEP 5

Apply the Fundamental Theorem of Calculus:
$\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x = F(a) - F(0)$
Substitute the antiderivative into the expression:
$F(a) = \frac{a^3}{27} + 3a^2$
$F(0) = \frac{0^3}{27} + 3(0)^2 = 0$

STEP 6

Evaluate the expression:
$F(a) - F(0) = \left(\frac{a^3}{27} + 3a^2\right) - 0$
Simplify:
$\frac{a^3}{27} + 3a^2$
The exact value of the integral is:
$\boxed{\frac{a^3}{27} + 3a^2}$

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QuestionEvaluate exactly, using the Fundamental Theorem of Calculus: ∫0a(x29+6x)dx=\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x=∫0a​(9x2​+6x)dx= □\square□

Studdy Solution

STEP 1

1. We are given the definite integral ∫0a(x29+6x)dx\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x∫0a​(9x2​+6x)dx.2. We will use the Fundamental Theorem of Calculus to evaluate the integral.

STEP 2

1. Identify the integrand and split the integral if necessary.2. Find the antiderivative of the integrand.3. Apply the Fundamental Theorem of Calculus to evaluate the definite integral.4. Simplify the result.

STEP 3

Identify the integrand:The integrand is x29+6x\frac{x^{2}}{9} + 6x9x2​+6x.

STEP 4

Find the antiderivative of the integrand:The antiderivative of x29\frac{x^{2}}{9}9x2​ is x327\frac{x^3}{27}27x3​.The antiderivative of 6x6x6x is 3x23x^23x2.Therefore, the antiderivative of the entire integrand is:F(x)=x327+3x2 F(x) = \frac{x^3}{27} + 3x^2 F(x)=27x3​+3x2

STEP 5

STEP 6

Evaluate the expression:F(a)−F(0)=(a327+3a2)−0 F(a) - F(0) = \left(\frac{a^3}{27} + 3a^2\right) - 0 F(a)−F(0)=(27a3​+3a2)−0Simplify:a327+3a2 \frac{a^3}{27} + 3a^2 27a3​+3a2 The exact value of the integral is:a327+3a2 \boxed{\frac{a^3}{27} + 3a^2} 27a3​+3a2​

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QuestionEvaluate exactly, using the Fundamental Theorem of Calculus: $\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x=$ $\square$

1. We are given the definite integral $\int_{0}^{a}\left(\frac{x^{2}}{9}+6 x\right) d x$ .
2. We will use the Fundamental Theorem of Calculus to evaluate the integral.

1. Identify the integrand and split the integral if necessary.
2. Find the antiderivative of the integrand.
3. Apply the Fundamental Theorem of Calculus to evaluate the definite integral.
4. Simplify the result.

Identify the integrand:
The integrand is $\frac{x^{2}}{9} + 6x$ .

Find the antiderivative of the integrand:
The antiderivative of $\frac{x^{2}}{9}$ is $\frac{x^3}{27}$ .
The antiderivative of $6x$ is $3x^2$ .
Therefore, the antiderivative of the entire integrand is:
$F(x) = \frac{x^3}{27} + 3x^2$

Evaluate the expression:
$F(a) - F(0) = \left(\frac{a^3}{27} + 3a^2\right) - 0$
Simplify:
$\frac{a^3}{27} + 3a^2$
The exact value of the integral is:
$\boxed{\frac{a^3}{27} + 3a^2}$