Evaluate int_(0)^(5)(7e^(0.2 x)+4x)dx and express the answer in simplest form. Answer:

Identify Integral: Identify the integral to be solved. We need to evaluate the integral of the function 7e^(0.2x) + 4x with respect to x from 0 to 5.Break into Two: Break the integral into two separate integrals. The integral of a sum is the sum of the integrals, so we can write: ∫(7e^(0.2x) + 4x)dx = ∫7e^(0.2x)dx + ∫4xdxEvaluate First Integral: Evaluate the first integral ∫7e^(0.2x)dx. To integrate 7e^(0.2x), we can use the substitution method or recognize that the integral of e^(ax) is (1/a)e^(ax), where a is a constant. So, ∫7e^(0.2x)dx = (7/0.2)e^(0.2x) = 35e^(0.2x)Evaluate Second Integral: Evaluate the second integral ∫4xdx. The integral of 4x with respect to x is 4x^2/2, which simplifies to 2x^2. So, ∫4xdx = 2x^2Combine Results: Combine the results of the two integrals. The combined integral is 35e^(0.2x) + 2x^2.Apply Limits: Apply the limits of integration from 0 to 5. We need to evaluate (35e^(0.2x) + 2x^2) from x = 0 to x = 5. So, we calculate (35e^(0.2*5) + 2*5^2) - (35e^(0.2*0) + 2*0^2).Perform Upper Limit: Perform the calculations for the upper limit x = 5. For x = 5, we have 35e^(0.2*5) + 2*5^2 = 35e^1 + 2*25 = 35e + 50.Perform Lower Limit: Perform the calculations for the lower limit x = 0. For x = 0, we have 35e^(0.2*0) + 2*0^2 = 35e^0 + 0 = 35*1 = 35.Subtract Values: Subtract the value at the lower limit from the value at the upper limit. (35e + 50) - 35 = 35e + 50 - 35 = 35e + 15.Write Final Answer: Write the final answer. The integral from 0 to 5 of the function 7e^(0.2x) + 4x is 35e + 15.

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Evaluate
int_(0)^(5)(7e^(0.2 x)+4x)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{5}\left(7 e^{0.2 x}+4 x\right) d x$ and express the answer in simplest form. $\newline$ Answer:

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Calculus

Evaluate definite integrals using the chain rule

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Q. Evaluate

\int_{0}^{5}\left(7 e^{0.2 x}+4 x\right) d x

and express the answer in simplest form.

\newline

Answer:

Identify Integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of the function $7e^{0.2x} + 4x$ with respect to $x$ from $0$ to $5$ .
Break into Two: Break the integral into two separate integrals. $\newline$ The integral of a sum is the sum of the integrals, so we can write: $\newline$ \int(\(7e^{ $0$ . $2$ x} + $4$ x)\,dx = \int $7$ e^{ $0$ . $2$ x}\,dx + \int $4$ x\,dx
Evaluate First Integral: Evaluate the first integral $\int 7e^{0.2x}\,dx$ . To integrate $7e^{0.2x}$ , we can use the substitution method or recognize that the integral of $e^{ax}$ is $(1/a)e^{ax}$ , where $a$ is a constant. So, $\int 7e^{0.2x}\,dx = (7/0.2)e^{0.2x} = 35e^{0.2x}$
Evaluate Second Integral: Evaluate the second integral $\int 4x\,dx$ . $\newline$ The integral of $4x$ with respect to $x$ is $\frac{4x^2}{2}$ , which simplifies to $2x^2$ . $\newline$ So, $\int 4x\,dx = 2x^2$
Combine Results: Combine the results of the two integrals. $\newline$ The combined integral is $35e^{0.2x} + 2x^2$ .
Apply Limits: Apply the limits of integration from $0$ to $5$ . We need to evaluate $(35e^{0.2x} + 2x^2)$ from $x = 0$ to $x = 5$ . So, we calculate $(35e^{0.2\times 5} + 2\times 5^2) - (35e^{0.2\times 0} + 2\times 0^2)$ .
Perform Upper Limit: Perform the calculations for the upper limit $x = 5$ . For $x = 5$ , we have $35e^{(0.2 \times 5)} + 2 \times 5^2 = 35e^1 + 2 \times 25 = 35e + 50$ .
Perform Lower Limit: Perform the calculations for the lower limit $x = 0$ . For $x = 0$ , we have $35e^{(0.2\cdot 0)} + 2\cdot 0^2 = 35e^0 + 0 = 35\cdot 1 = 35$ .
Subtract Values: Subtract the value at the lower limit from the value at the upper limit. $\newline$ $(35e + 50) - 35 = 35e + 50 - 35 = 35e + 15$ .
Write Final Answer: Write the final answer. $\newline$ The integral from $0$ to $5$ of the function $7e^{0.2x} + 4x$ is $35e + 15$ .