evaluate the integral.\n∫te^5t dt\nstep 1\nrecall the formula for integration by parts, which states that if…
evaluate the integral.\n∫te^5t dt\nstep 1\nrecall the formula for integration by parts, which states that if f and g are differentiable functions, then the following holds.\n∫f(x)g(x) dx = f(x)g(x) - ∫g(x)f(x) dx\nif we let u = f(x) and v = g(x), then the differentials are f(x) dx and dv = g(x) dx. so, by the substitution rule, the formula for integration by parts becomes the following.\n∫u dv = uv - ∫v du\nwe are given ∫te^5t dt. in order to use the formula for integration by parts, we can choose to let u = t and dv = (e^5t) dt.\nstep 2\nif u = t and dv = e^5t dt, then we have the following.\ndu = dt\nv = (e^5t) ×
Answer
Explanation:
Step1: Recall integration - by - parts formula
The formula for integration by parts is $\int u\mathrm{d}v=uv - \int v\mathrm{d}u$. For the integral $\int te^{5t}\mathrm{d}t$, we choose $u = t$ and $\mathrm{d}v=e^{5t}\mathrm{d}t$.
Step2: Find $\mathrm{d}u$ and $v$
If $u = t$, then $\mathrm{d}u=\mathrm{d}t$. To find $v$ from $\mathrm{d}v = e^{5t}\mathrm{d}t$, we integrate $\mathrm{d}v$ with respect to $t$. Using the rule $\int e^{ax}\mathrm{d}x=\frac{1}{a}e^{ax}+C$ ($a = 5$ here), we have $v=\frac{1}{5}e^{5t}$.
Step3: Apply the integration - by - parts formula
Substitute $u = t$, $\mathrm{d}u=\mathrm{d}t$, $v=\frac{1}{5}e^{5t}$ and $\mathrm{d}v=e^{5t}\mathrm{d}t$ into $\int u\mathrm{d}v=uv - \int v\mathrm{d}u$. We get $\int te^{5t}\mathrm{d}t=t\times\frac{1}{5}e^{5t}-\int\frac{1}{5}e^{5t}\mathrm{d}t$.
Step4: Evaluate the remaining integral
$\int\frac{1}{5}e^{5t}\mathrm{d}t=\frac{1}{25}e^{5t}+C$. So $\int te^{5t}\mathrm{d}t=\frac{1}{5}te^{5t}-\frac{1}{25}e^{5t}+C=\frac{e^{5t}(5t - 1)}{25}+C$.
Answer:
$\frac{e^{5t}(5t - 1)}{25}+C$