Answer to Question #212551 in Calculus for Akhona Klanisi

Question

Evaluate the definite intergral $\int$₀¹ 18e^3x+1dx (round to an interger).

1.         322

2.         311

3.         932

4.             965

Answer 1

Let,

$3x+1=t$

differentiating above equation with respect to x

$3dx=dt$

$dx=\dfrac{dt}{3}$

When $x=0;\space t=1$

$x=1;\space t=4$

$∫_0^1 18e^{3x+1}dx$

Substituting $(3x+1=t)$ in above problem and changing limits of integration

$= ∫_1^4 \dfrac{18e^{t}dt}{3}$

$=6∫_1^4 e^{t}dt$

$=6 |e^{t}|_1^4$

$=6(e^4-e)$

$=311.27\approx311$