Evaluate int_(0)^(4)(8e^(-0.5 x)-2x)dx and express the answer in simplest form. Answer:

Break down integral: Break down the integral into two separate integrals. We have the integral of a sum of two functions, which can be separated into the sum of two integrals: ∫(8e^(-0.5x) - 2x)dx = ∫8e^(-0.5x)dx - ∫2xdxEvaluate first integral: Evaluate the first integral ∫8e^(-0.5x)dx. To integrate 8e^(-0.5x), we use the substitution method: Let u = -0.5x, then du = -0.5dx, which means dx = -2du. The integral becomes: ∫8e^u * -2du = -16∫e^u du The integral of e^u with respect to u is e^u, so we have: -16e^u + C Substitute back u = -0.5x to get: -16e^(-0.5x) + CEvaluate second integral: Evaluate the second integral ∫2xdx. The integral of 2x with respect to x is x^2, so we have: ∫2xdx = 2 * (1/2)x^2 = x^2 + CCombine results: Combine the results from Step 2 and Step 3 to get the indefinite integral. The indefinite integral is: -16e^(-0.5x) + x^2 + CEvaluate definite integral: Evaluate the definite integral from 0 to 4. We need to calculate the value of the indefinite integral at the upper limit (x=4) and subtract the value at the lower limit (x=0): (-16e^(-0.5*4) + 4^2) - (-16e^(-0.5*0) + 0^2) Simplify and calculate the values: (-16e^(-2) + 16) - (-16e^(0) + 0) (-16/e^2 + 16) - (-16 + 0) -16/e^2 + 16 + 16 16 + 16 - 16/e^2 32 - 16/e^2

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Evaluate
int_(0)^(4)(8e^(-0.5 x)-2x)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{4}\left(8 e^{-0.5 x}-2 x\right) d x$ and express the answer in simplest form. $\newline$ Answer:

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Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Evaluate

\int_{0}^{4}\left(8 e^{-0.5 x}-2 x\right) d x

and express the answer in simplest form.

\newline

Answer:

Break down integral: Break down the integral into two separate integrals. $\newline$ We have the integral of a sum of two functions, which can be separated into the sum of two integrals: $\newline$ \int(\(8e^{ $-0$ . $5$ x} - $2$ x)\,dx = \int $8$ e^{ $-0$ . $5$ x}\,dx - \int $2$ x\,dx
Evaluate first integral: Evaluate the first integral $\int 8e^{-0.5x}\,dx$ . To integrate $8e^{-0.5x}$ , we use the substitution method: Let $u = -0.5x$ , then $du = -0.5dx$ , which means $dx = -2du$ . The integral becomes: $\int 8e^u \cdot -2du = -16\int e^u \,du$ The integral of $e^u$ with respect to $u$ is $e^u$ , so we have: $-16e^u + C$ Substitute back $u = -0.5x$ to get: $8e^{-0.5x}$ $1$
Evaluate second integral: Evaluate the second integral $\int 2x \, dx$ . $\newline$ The integral of $2x$ with respect to $x$ is $x^2$ , so we have: $\newline$ $\int 2x \, dx = 2 \cdot (\frac{1}{2})x^2 = x^2 + C$
Combine results: Combine the results from Step $2$ and Step $3$ to get the indefinite integral. $\newline$ The indefinite integral is: $\newline$ $-16e^{(-0.5x)} + x^2 + C$
Evaluate definite integral: Evaluate the definite integral from $0$ to $4$ . We need to calculate the value of the indefinite integral at the upper limit $(x=4)$ and subtract the value at the lower limit $(x=0)$ : $(-16e^{(-0.5\cdot4)} + 4^2) - (-16e^{(-0.5\cdot0)} + 0^2)$ Simplify and calculate the values: $(-16e^{(-2)} + 16) - (-16e^{(0)} + 0)$ $(-16/e^2 + 16) - (-16 + 0)$ $-16/e^2 + 16 + 16$ $16 + 16 - 16/e^2$ $32 - 16/e^2$