Question

The value of the integral \(\int_0^1(x+3) \,e^{3x} \,dx\).

(a) \(\frac{8e^3}{9}\)

(b) \(\frac{11}{9} e^3-8\)

(c) \(\frac{e^{3x}}{9}(x+8)\)

(d) \(\frac{11}{9} e^3-\frac{8}{9}\)

I had been asked this question at a job interview.

My question comes from Fundamental Theorem of Calculus-2 in chapter Integrals of Mathematics – Class 12

Answer 1

Correct choice is (d) \(\frac{11}{9} e^3-\frac{8}{9}\)

The explanation: Let \(I=\int_0^1(x+3) \,e^{3x} \,dx\)

F(x)=\(\int (x+3) \,e^{3x} \,dx\)

By using the formula \(\int \,u.v \,dx=u\int \,v dx-\int \,u'(\int \,v \,dx)\), we get

F(x)=(x+3) \(\int e^{3x} \,dx-\int \,(x+3)’\int \,e^{3x} \,dx\)

=\(\frac{(x+3) \,e^{3x}}{3}-\int \frac{e^{3x}}{3} dx\)

=\(\frac{(x+3) e^{3x}}{3}-\frac{e^{3x}}{9}\)

=\(\frac{e^{3x}}{3} (x+3-\frac{1}{3})=\frac{e^{3x}}{9}(3x+8)\)

Applying the limits, we get

I=F(1)-F(0)

=\(\frac{e^{3(1)}}{9} (3+8)-\frac{e^{3(0)}}{9}(0+8)\)

=\(\frac{11}{9} e^3-\frac{8}{9}\).