Evaluate int_(0)^(4)(2e^(-0.25 x)-4x)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: ∫(2e^(-0.25x) - 4x)dx = ∫2e^(-0.25x)dx - ∫4xdxEvaluate first integral: Evaluate the first integral ∫2e^(-0.25x)dx. To integrate 2e^(-0.25x), we use the substitution method: Let u = -0.25x, then du = -0.25dx, or dx = -4du. The limits of integration also change with the substitution. When x = 0, u = 0, and when x = 4, u = -1. The integral becomes: ∫2e^u * -4du = -8∫e^u du The integral of e^u with respect to u is e^u, so we have: -8e^u + C Substituting back for u, we get: -8e^(-0.25x) + CEvaluate second integral: Evaluate the second integral ∫4xdx. The integral of 4x with respect to x is 2x^2, so we have: 2x^2 + CCombine results: Combine the results of the two integrals. The combined indefinite integral is: -8e^(-0.25x) + 2x^2 + CEvaluate definite integral: Evaluate the definite integral from 0 to 4. We need to calculate the value of the combined integral at the upper limit (x = 4) and subtract the value at the lower limit (x = 0): [-8e^(-0.25*4) + 2*4^2] - [-8e^(-0.25*0) + 2*0^2] Calculating the values, we get: [-8e^(-1) + 2*16] - [-8e^(0) + 0] [-8/e + 32] - [-8*1] -8/e + 32 + 8Simplify final result: Simplify the final result. Combining the terms, we get: 32 + 8 - 8/e 32 + 8(1 - 1/e) 40 - 8/e

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Evaluate
int_(0)^(4)(2e^(-0.25 x)-4x)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{4}\left(2 e^{-0.25 x}-4 x\right) d x$ and express the answer in simplest form. $\newline$ Answer:

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Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Evaluate

\int_{0}^{4}\left(2 e^{-0.25 x}-4 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(\(2e^{ $-0$ . $25$ x} - $4$ x)\,dx = \int $2$ e^{ $-0$ . $25$ x}\,dx - \int $4$ x\,dx
Evaluate first integral: Evaluate the first integral $\int 2e^{-0.25x}\,dx$ . To integrate $2e^{-0.25x}$ , we use the substitution method: Let $u = -0.25x$ , then $du = -0.25dx$ , or $dx = -4du$ . The limits of integration also change with the substitution. When $x = 0$ , $u = 0$ , and when $x = 4$ , $u = -1$ . The integral becomes: $\int 2e^u \cdot -4du = -8\int e^u \,du$ The integral of $2e^{-0.25x}$ $0$ with respect to $2e^{-0.25x}$ $1$ is $2e^{-0.25x}$ $0$ , so we have: $2e^{-0.25x}$ $3$ Substituting back for $2e^{-0.25x}$ $1$ , we get: $2e^{-0.25x}$ $5$
Evaluate second integral: Evaluate the second integral $\int 4x\,dx$ . The integral of $4x$ with respect to $x$ is $2x^2$ , so we have: $2x^2 + C$
Combine results: Combine the results of the two integrals. $\newline$ The combined indefinite integral is: $\newline$ $-8e^{(-0.25x)} + 2x^2 + C$
Evaluate definite integral: Evaluate the definite integral from $0$ to $4$ . We need to calculate the value of the combined integral at the upper limit ( $x = 4$ ) and subtract the value at the lower limit ( $x = 0$ ): $[-8e^{(-0.25\cdot4)} + 2\cdot4^2] - [-8e^{(-0.25\cdot0)} + 2\cdot0^2]$ Calculating the values, we get: $[-8e^{(-1)} + 2\cdot16] - [-8e^{(0)} + 0]$ $[-\frac{8}{e} + 32] - [-8\cdot1]$ $-\frac{8}{e} + 32 + 8$
Simplify final result: Simplify the final result. $\newline$ Combining the terms, we get: $\newline$ $32 + 8 - \frac{8}{e}$ $\newline$ $32 + 8(1 - \frac{1}{e})$ $\newline$ $40 - \frac{8}{e}$