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Math

Math Snap

PROBLEM

Use the Fundamental Theorem to evaluate the definite integral exactly.
03(12x2+1)dx\int_{0}^{3}\left(12 x^{2}+1\right) d x Enter the exact answer.
03(12x2+1)dx=\int_{0}^{3}\left(12 x^{2}+1\right) d x= \square
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STEP 1

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

STEP 4

Evaluate the antiderivative F(x) F(x) at the upper limit x=3 x = 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)=108+3 F(3) = 108 + 3 F(3)=111 F(3) = 111

STEP 5

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

SOLUTION

Subtract the value of the antiderivative at the lower limit from the value at the upper limit:
03(12x2+1)dx=F(3)F(0) \int_{0}^{3}(12x^2 + 1) \, dx = F(3) - F(0) 03(12x2+1)dx=1110 \int_{0}^{3}(12x^2 + 1) \, dx = 111 - 0 03(12x2+1)dx=111 \int_{0}^{3}(12x^2 + 1) \, dx = 111 The exact value of the definite integral is:
111 \boxed{111}

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