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

Category: QuestionsThe value of the integral \int_0^1(x+3) \,e^{3x} \,dx.
Editor">Editor Staff asked 11 months ago

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}
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My question comes from Fundamental Theorem of Calculus-2 in chapter Integrals of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

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
=\frac{e^{3(1)}}{9} (3+8)-\frac{e^{3(0)}}{9}(0+8)
=\frac{11}{9} e^3-\frac{8}{9}.