Math / Calculus

QuestionUse the Fundamental Theorem to evaluate the definite integral exactly. $\int_{0}^{3}\left(12 x^{2}+1\right) d x$
Enter the exact answer. $\int_{0}^{3}\left(12 x^{2}+1\right) d x=$ $\square$ eTextbook and Media

Studdy Solution

STEP 1

1. The function to integrate is $f(x) = 12x^2 + 1$ .
2. We will use the Fundamental Theorem of Calculus to evaluate the definite integral.
3. The limits of integration are from 0 to 3.

STEP 2

1. Find the antiderivative of the function $f(x) = 12x^2 + 1$ .
2. Evaluate the antiderivative at the upper and lower limits of integration.
3. Subtract the value of the antiderivative at the lower limit from the value at the upper limit to find the exact value of the definite integral.

STEP 3

Find the antiderivative of $f(x) = 12x^2 + 1$ .
The antiderivative of $12x^2$ is $\frac{12}{3}x^3 = 4x^3$ .
The antiderivative of $1$ is $x$ .
Thus, the antiderivative $F(x)$ of $f(x)$ is:
$F(x) = 4x^3 + x$

STEP 4

Evaluate the antiderivative $F(x)$ at the upper limit $x = 3$ :
$F(3) = 4(3)^3 + 3$ $F(3) = 4 \times 27 + 3$ $F(3) = 108 + 3$ $F(3) = 111$

STEP 5

Evaluate the antiderivative $F(x)$ at the lower limit $x = 0$ :
$F(0) = 4(0)^3 + 0$ $F(0) = 0$

STEP 6

Subtract the value of the antiderivative at the lower limit from the value at the upper limit:
$\int_{0}^{3}(12x^2 + 1) \, dx = F(3) - F(0)$ $\int_{0}^{3}(12x^2 + 1) \, dx = 111 - 0$ $\int_{0}^{3}(12x^2 + 1) \, dx = 111$
The exact value of the definite integral is:
$\boxed{111}$

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QuestionUse the Fundamental Theorem to evaluate the definite integral exactly. $\int_{0}^{3}\left(12 x^{2}+1\right) d x$
Enter the exact answer. $\int_{0}^{3}\left(12 x^{2}+1\right) d x=$ $\square$ eTextbook and Media

Studdy Solution

STEP 1

1. The function to integrate is $f(x) = 12x^2 + 1$ .
2. We will use the Fundamental Theorem of Calculus to evaluate the definite integral.
3. The limits of integration are from 0 to 3.

STEP 2

1. Find the antiderivative of the function $f(x) = 12x^2 + 1$ .
2. Evaluate the antiderivative at the upper and lower limits of integration.
3. Subtract the value of the antiderivative at the lower limit from the value at the upper limit to find the exact value of the definite integral.

STEP 3

Find the antiderivative of $f(x) = 12x^2 + 1$ .
The antiderivative of $12x^2$ is $\frac{12}{3}x^3 = 4x^3$ .
The antiderivative of $1$ is $x$ .
Thus, the antiderivative $F(x)$ of $f(x)$ is:
$F(x) = 4x^3 + x$

STEP 4

Evaluate the antiderivative $F(x)$ at the upper limit $x = 3$ :
$F(3) = 4(3)^3 + 3$ $F(3) = 4 \times 27 + 3$ $F(3) = 108 + 3$ $F(3) = 111$

STEP 5

Evaluate the antiderivative $F(x)$ at the lower limit $x = 0$ :
$F(0) = 4(0)^3 + 0$ $F(0) = 0$

STEP 6

Subtract the value of the antiderivative at the lower limit from the value at the upper limit:
$\int_{0}^{3}(12x^2 + 1) \, dx = F(3) - F(0)$ $\int_{0}^{3}(12x^2 + 1) \, dx = 111 - 0$ $\int_{0}^{3}(12x^2 + 1) \, dx = 111$
The exact value of the definite integral is:
$\boxed{111}$

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QuestionUse the Fundamental Theorem to evaluate the definite integral exactly. ∫03(12x2+1)dx\int_{0}^{3}\left(12 x^{2}+1\right) d x∫03​(12x2+1)dxEnter the exact answer. ∫03(12x2+1)dx=\int_{0}^{3}\left(12 x^{2}+1\right) d x=∫03​(12x2+1)dx= □\square□ eTextbook and Media

Studdy Solution

STEP 1

1. The function to integrate is f(x)=12x2+1 f(x) = 12x^2 + 1 f(x)=12x2+1.2. We will use the Fundamental Theorem of Calculus to evaluate the definite integral.3. The limits of integration are from 0 to 3.

STEP 2

STEP 3

Find the antiderivative of f(x)=12x2+1 f(x) = 12x^2 + 1 f(x)=12x2+1.The antiderivative of 12x2 12x^2 12x2 is 123x3=4x3 \frac{12}{3}x^3 = 4x^3 312​x3=4x3.The antiderivative of 1 1 1 is x x x.Thus, the antiderivative F(x) F(x) F(x) of f(x) f(x) f(x) is:F(x)=4x3+x F(x) = 4x^3 + x F(x)=4x3+x

STEP 4

Evaluate the antiderivative F(x) F(x) F(x) at the upper limit x=3 x = 3 x=3:F(3)=4(3)3+3 F(3) = 4(3)^3 + 3 F(3)=4(3)3+3 F(3)=4×27+3 F(3) = 4 \times 27 + 3 F(3)=4×27+3 F(3)=108+3 F(3) = 108 + 3 F(3)=108+3 F(3)=111 F(3) = 111 F(3)=111

STEP 5

Evaluate the antiderivative F(x) F(x) F(x) at the lower limit x=0 x = 0 x=0:F(0)=4(0)3+0 F(0) = 4(0)^3 + 0 F(0)=4(0)3+0 F(0)=0 F(0) = 0 F(0)=0

STEP 6

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QuestionUse the Fundamental Theorem to evaluate the definite integral exactly. $\int_{0}^{3}\left(12 x^{2}+1\right) d x$
Enter the exact answer. $\int_{0}^{3}\left(12 x^{2}+1\right) d x=$ $\square$ eTextbook and Media

1. The function to integrate is $f(x) = 12x^2 + 1$ .
2. We will use the Fundamental Theorem of Calculus to evaluate the definite integral.
3. The limits of integration are from 0 to 3.

Find the antiderivative of $f(x) = 12x^2 + 1$ .
The antiderivative of $12x^2$ is $\frac{12}{3}x^3 = 4x^3$ .
The antiderivative of $1$ is $x$ .
Thus, the antiderivative $F(x)$ of $f(x)$ is:
$F(x) = 4x^3 + x$

Evaluate the antiderivative $F(x)$ at the upper limit $x = 3$ :
$F(3) = 4(3)^3 + 3$ $F(3) = 4 \times 27 + 3$ $F(3) = 108 + 3$ $F(3) = 111$

Evaluate the antiderivative $F(x)$ at the lower limit $x = 0$ :
$F(0) = 4(0)^3 + 0$ $F(0) = 0$