int {frac{t}{e^{3t^2}}} , dt = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I = \int t e^{-3t^2} \, dt$. Substitute $z = -3t^2$,then $dz = -6t \, dt$,which implies $t \, dt = -\frac{1}{6} \, dz$. Substituting these in

Vedclass · Accepted Answer

$-\frac{1}{6} e^{-3t^2} + c$

Vedclass · Answer

(D) Let $I = \int t e^{-3t^2} \, dt$. Substitute $z = -3t^2$,then $dz = -6t \, dt$,which implies $t \, dt = -\frac{1}{6} \, dz$. Substituting these into the integral: $I = \int e^z \left(-\frac{1}{6}\right) \, dz = -\frac{1}{6} \int e^z \, dz$. Integrating $e^z$ gives $e^z$: $I = -\frac{1}{6} e^z + c$. Substituting back $z = -3t^2$: $I = -\frac{1}{6} e^{-3t^2} + c$.

$\int {\frac{t}{e^{3t^2}}} \, dt = $

Similar Questions

Integrate the function: $\frac{\sin x}{\sin (x-a)}$

If $\int \frac{\log _e(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx = f(g(x)) + c$, then:

$\int \frac{1}{(2 \cos x+\sin x)^2} d x=$

$\int {\sqrt {{e^x} - 1} } dx = $

Integrate the function $\frac{x}{9-4x^{2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module