Evaluate int_(0)^(4)(5e^(-0.25 x)-4x)dx and express the answer in simplest form. Answer:

Break down into two integrals: 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: ∫(5e^(-0.25x) - 4x)dx = ∫5e^(-0.25x)dx - ∫4xdxEvaluate first integral: Evaluate the first integral ∫5e^(-0.25x)dx. To integrate 5e^(-0.25x), we can use the substitution method: Let u = -0.25x, then du = -0.25dx, which means 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: ∫5e^u(-4du) = -20∫e^udu The integral of e^u with respect to u is e^u, so we have: -20e^u + C Substituting back for u, we get: -20e^(-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 from Step 2 and Step 3. The combined integral from 0 to 4 is: -20e^(-0.25x) + 2x^2 evaluated from 0 to 4.Evaluate at upper limit: Evaluate the combined integral at the upper limit x = 4. -20e^(-0.25*4) + 2*4^2 = -20e^(-1) + 2*16 = -20/e + 32Evaluate at lower limit: Evaluate the combined integral at the lower limit x = 0. -20e^(-0.25*0) + 2*0^2 = -20e^0 + 0 = -20Subtract values: Subtract the value of the integral at the lower limit from the value at the upper limit. (-20/e + 32) - (-20) = -20/e + 32 + 20 = -20/e + 52

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Evaluate
int_(0)^(4)(5e^(-0.25 x)-4x)dx and express the answer in simplest form.
Answer:

Evaluate $\int_{0}^{4}\left(5 e^{-0.25 x}-4 x\right) d x$ 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(5 e^{-0.25 x}-4 x\right) d x

and express the answer in simplest form.

\newline

Answer:

Break down into two integrals: 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(\(5e^{ $-0$ . $25$ x} - $4$ x)\,dx = \int $5$ e^{ $-0$ . $25$ x}\,dx - \int $4$ x\,dx
Evaluate first integral: Evaluate the first integral $\int 5e^{-0.25x}\,dx$ . To integrate $5e^{-0.25x}$ , we can use the substitution method: Let $u = -0.25x$ , then $du = -0.25dx$ , which means $dx = -4du$ . The limits of integration also change with the substitution. When $x = 0$ , $u = 0$ , and when $x = 4$ , $u = -1$ . The integral becomes: $\int 5e^u(-4du) = -20\int e^u\,du$ The integral of $5e^{-0.25x}$ $0$ with respect to $5e^{-0.25x}$ $1$ is $5e^{-0.25x}$ $0$ , so we have: $5e^{-0.25x}$ $3$ Substituting back for $5e^{-0.25x}$ $1$ , we get: $5e^{-0.25x}$ $5$
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 from Step $2$ and Step $3$ . $\newline$ The combined integral from $0$ to $4$ is: $\newline$ $-20e^{(-0.25x)} + 2x^2$ evaluated from $0$ to $4$ .
Evaluate at upper limit: Evaluate the combined integral at the upper limit $x = 4$ .
$-20e^{(-0.25\cdot 4)} + 2\cdot 4^2$
= $-20e^{(-1)} + 2\cdot 16$
= $-\frac{20}{e} + 32$
Evaluate at lower limit: Evaluate the combined integral at the lower limit $x = 0$ . $-20e^{(-0.25\cdot 0)} + 2\cdot 0^2 = -20e^0 + 0 = -20$
Subtract values: Subtract the value of the integral at the lower limit from the value at the upper limit. $\newline$ $(-20/e + 32) - (-20)$ $\newline$ $= -20/e + 32 + 20$ $\newline$ $= -20/e + 52$