Question

Given that \(\int_0^1 3 x\sqrt {{x^2} + 4} dx = 5\sqrt 5 - 8\), what is \(\int_1^0 3 u\sqrt {{u^2} + 4} du?\)

Short answer

The value of \(\int_1^0 3 u\sqrt {{u^2} + 4} du\) is \( - 5\sqrt 5 + 8\)

Step-by-step solution

Property used

The definite integral, \(\int_a^b f (x)dx\), in which the limits are \(a\) and \(b\).

The length of the subinterval is\(\Delta x = \frac{{b - a}}{n}\).

When the limits\(a\)and\(b\)are reversed,\(\Delta x\)changes from\(\frac{{(b - a)}}{n}\)to\(\frac{{(a - b)}}{n}\).

Hence, \(\int_b^a f (x)dx = - \int_a^b f (x)dx\)

Calculation of the integral

The value of the integral\(\int_0 3 x\sqrt {{x^2} + 4} dx\)is\(5\sqrt 5 - 8\)

Obtain the value of the integral\(\int_1^0 3 u\sqrt {{u^2} + 4} du\)as follows.

It is noted from the property of the integrals that,\(\int_1^0 3 u\sqrt {{u^2} + 4} du = - \int_0^1 3 u\sqrt {{u^2} + 4} du\).

Comparing the integrals\( - \int_0^1 3 u\sqrt {{u^2} + 4} du\)and\(\int_0^1 3 x\sqrt {{x^2} + 4} dx\), It is discovered that both integrals have the same form, with a negative sign in one integral and limits ranging from 0 to 1.

So, \( - \int_0^1 3 u\sqrt {{u^2} + 4} du = - (5\sqrt 5 - 8).\)

Evaluation of the integral

Hence, the value of the integral\(\int_1^0 3 u\sqrt {{u^2} + 4} \)is evaluated as,

\(\int_1^0 3 u\sqrt {{u^2} + 4} du = - \int_0^1 3 u\sqrt {{u^2} + 4} du\)

\( = - \int_0^1 3 x\sqrt {{x^2} + 4} dx\)

\( = - (5\sqrt 5 - 8)\)

\( = - 5\sqrt 5 + 8\)

Hence, the value of \(\int_1^0 3 u\sqrt {{u^2} + 4} du\) is \( - 5\sqrt 5 + 8\).