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Sagot :
Sure, let's go through the steps to evaluate the definite integral [tex]\(\int_0^5(4t - 1)^2 \, dt\)[/tex].
### Step 1: Expand the Integrand
First, let’s expand the integrand [tex]\((4t - 1)^2\)[/tex]:
[tex]\[ (4t - 1)^2 = (4t - 1)(4t - 1) = 16t^2 - 8t + 1 \][/tex]
### Step 2: Set Up the Integral
Now, rewrite the integral with the expanded form:
[tex]\[ \int_0^5 (16t^2 - 8t + 1) \, dt \][/tex]
### Step 3: Integrate Term by Term
Next, we will integrate each term separately.
1. The integral of [tex]\(16t^2\)[/tex]:
[tex]\[ \int 16t^2 \, dt = 16 \cdot \frac{t^3}{3} = \frac{16}{3}t^3 \][/tex]
2. The integral of [tex]\(-8t\)[/tex]:
[tex]\[ \int -8t \, dt = -8 \cdot \frac{t^2}{2} = -4t^2 \][/tex]
3. The integral of [tex]\(1\)[/tex]:
[tex]\[ \int 1 \, dt = t \][/tex]
### Step 4: Combine the Integrals
Now, combine these results:
[tex]\[ \int_0^5 (16t^2 - 8t + 1) \, dt = \left[ \frac{16}{3} t^3 - 4t^2 + t \right]_0^5 \][/tex]
### Step 5: Evaluate the Definite Integral
Now evaluate the antiderivative from [tex]\(0\)[/tex] to [tex]\(5\)[/tex].
First, plug in the upper limit [tex]\(t = 5\)[/tex]:
[tex]\[ \left( \frac{16}{3} \cdot 5^3 - 4 \cdot 5^2 + 5 \right) = \frac{16}{3} \cdot 125 - 4 \cdot 25 + 5 = \frac{2000}{3} - 100 + 5 \][/tex]
Simplify:
[tex]\[ \frac{2000}{3} - 95 = \frac{2000}{3} - \frac{285}{3} = \frac{2000 - 285}{3} = \frac{1715}{3} \][/tex]
Next, plug in the lower limit [tex]\(t = 0\)[/tex]:
[tex]\[ \left( \frac{16}{3} \cdot 0^3 - 4 \cdot 0^2 + 0 \right) = 0 \][/tex]
Subtract the results:
[tex]\[ \frac{1715}{3} - 0 = \frac{1715}{3} \][/tex]
### Final Answer
Thus, the value of the definite integral is:
[tex]\[ \boxed{\frac{1715}{3}} \][/tex]
### Step 1: Expand the Integrand
First, let’s expand the integrand [tex]\((4t - 1)^2\)[/tex]:
[tex]\[ (4t - 1)^2 = (4t - 1)(4t - 1) = 16t^2 - 8t + 1 \][/tex]
### Step 2: Set Up the Integral
Now, rewrite the integral with the expanded form:
[tex]\[ \int_0^5 (16t^2 - 8t + 1) \, dt \][/tex]
### Step 3: Integrate Term by Term
Next, we will integrate each term separately.
1. The integral of [tex]\(16t^2\)[/tex]:
[tex]\[ \int 16t^2 \, dt = 16 \cdot \frac{t^3}{3} = \frac{16}{3}t^3 \][/tex]
2. The integral of [tex]\(-8t\)[/tex]:
[tex]\[ \int -8t \, dt = -8 \cdot \frac{t^2}{2} = -4t^2 \][/tex]
3. The integral of [tex]\(1\)[/tex]:
[tex]\[ \int 1 \, dt = t \][/tex]
### Step 4: Combine the Integrals
Now, combine these results:
[tex]\[ \int_0^5 (16t^2 - 8t + 1) \, dt = \left[ \frac{16}{3} t^3 - 4t^2 + t \right]_0^5 \][/tex]
### Step 5: Evaluate the Definite Integral
Now evaluate the antiderivative from [tex]\(0\)[/tex] to [tex]\(5\)[/tex].
First, plug in the upper limit [tex]\(t = 5\)[/tex]:
[tex]\[ \left( \frac{16}{3} \cdot 5^3 - 4 \cdot 5^2 + 5 \right) = \frac{16}{3} \cdot 125 - 4 \cdot 25 + 5 = \frac{2000}{3} - 100 + 5 \][/tex]
Simplify:
[tex]\[ \frac{2000}{3} - 95 = \frac{2000}{3} - \frac{285}{3} = \frac{2000 - 285}{3} = \frac{1715}{3} \][/tex]
Next, plug in the lower limit [tex]\(t = 0\)[/tex]:
[tex]\[ \left( \frac{16}{3} \cdot 0^3 - 4 \cdot 0^2 + 0 \right) = 0 \][/tex]
Subtract the results:
[tex]\[ \frac{1715}{3} - 0 = \frac{1715}{3} \][/tex]
### Final Answer
Thus, the value of the definite integral is:
[tex]\[ \boxed{\frac{1715}{3}} \][/tex]
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