Question

The integral value of \(\int_0^a \int_0^x \int_0^{x+y} e^{x+y+z} \,dz \,dy \,dx\) is given by _____

(a) \(=\frac{1}{3}(e^{4a}+6e^{2a}+8e^{a}+3)\)

(b) \(=\frac{1}{3}(e^{4a}-6e^{2a}+4e^{a}+3)\)

(c) \(=\frac{1}{8}(e^{4a}-6e^{2a}+8e^{a}-3)\)

(d) 0

This question was addressed to me during an internship interview.

The doubt is from Triple Integral in division Multiple Integrals of Engineering Mathematics

Answer 1

Right answer is (c) \(=\frac{1}{8}(e^{4a}-6e^{2a}+8e^{a}-3)\)

Best explanation: \(\int_0^a \int_0^x \int_0^{x+y} e^{x+y+z} \,dz \,dy \,dx = \int_0^a \int_0^x \int_0^{x+y} e^{x+y} e^{z} \,dz \,dy \,dx\)

\(\int_0^a \int_0^x e^{x+y} [e^{z}]_0^{x+y} \,dy \,dx = \int_0^a \int_0^x e^{x+y} (e^{x+y}-1) \,dy \,dx\)

\(\int_0^a \int_0^x(e^{2x+2y}-e^{x+y}) \,dy \,dx = \int_0^a {e^{2x} \big[\frac{e^{2y}}{2}\big]_0^x – e^x [e^y]_0^x} \,dx\)

\(\int_0^a(\frac{e^{4x}}{2} – \frac{3}{2} e^{2x} + e^x)dx = \big[\frac{e^{4x}}{8} – \frac{3}{4} e^{2x} + e^x \big]_0^a \)

\(= (\frac{e^{4a}}{8} – \frac{3}{4} e^{2a} + e^a)-(\frac{1}{8}-\frac{3}{4}+1)\)

\(=\frac{1}{8}(e^{4a}-6e^{2a}+8e^{a}-3)\).