Calc 2 Problem. Could someone help me understand the "u" notation in this integral?

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Calc 2 Problem. Could someone help me understand the "u" notation in this integral?

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Answer 1 · 2018-04-14T02:58:56

$\int_{1}^{0} 12 u \sqrt{u^{2} + 1} d u = - (8 \sqrt{2} - 4) = 4 - 8 \sqrt{2}$ $int_1^0 12usqrt(u^2+1)du=-(8sqrt2-4)=4-8sqrt2$

Explanation:

The two provided integrals look almost exactly the same. In the first, the integrand is in terms of $x,$ $x,$ in the latter, the integrand has all $x$ $x$ replaced with $u$ $u$ and is otherwise the same.

The bounds have flipped between the two integrals. Fortunately, there is a rule for handling this:

$\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ $int_a^bf(x)dx=-int_b^af(x)dx$

Well, since we know $\int_{0}^{1} 12 x \sqrt{x^{2} + 1} d x = 8 \sqrt{2} - 4$ $int_0^1 12xsqrt(x^2+1)dx=8sqrt2-4$ , then, for the integral with flipped bounds and a similar integrand, we only have the same answer, but negative:

$\int_{1}^{0} 12 u \sqrt{u^{2} + 1} d u = - (8 \sqrt{2} - 4) = 4 - 8 \sqrt{2}$ $int_1^0 12usqrt(u^2+1)du=-(8sqrt2-4)=4-8sqrt2$

Answer 2 · 2018-04-14T03:44:10

The use of $u$ $u$ is irrelevant.

Explanation:

The definite integral is a number. There is really no variable.

We are given

With $x$ $x$ as the "dummy variable"

$\int_{0}^{1} 12 x \sqrt{x^{2} + 1} d x = 8 \sqrt{2} - 4$ $int_0^1 12xsqrt(x^2+1) dx = 8sqrt2-4$

and also with $t$ $t$ as the "dummy variable"

$\int_{0}^{1} 12 t \sqrt{t^{2} + 1} d t = 8 \sqrt{2} - 4$ $int_0^1 12tsqrt(t^2+1) dt = 8sqrt2-4$

and with $u$ $u$ as the "dummy variable"

$\int_{0}^{1} 12 u \sqrt{u^{2} + 1} d u = 8 \sqrt{2} - 4$ $int_0^1 12usqrt(u^2+1) du = 8sqrt2-4$

and so on.

The question asks us to find

$\int_{1}^{0} 12 u \sqrt{u^{2} + 1} d u$ $int_1^0 12usqrt(u^2+1) du$

which is exactly the same as

$\int_{1}^{0} 12 x \sqrt{x^{2} + 1} d x$ $int_1^0 12xsqrt(x^2+1) dx$

and

$\int_{1}^{0} 12 t \sqrt{t^{2} + 1} d t$ $int_1^0 12tsqrt(t^2+1) dt$

and

$\int_{1}^{0} 12 r \sqrt{r^{2} + 1} d r$ $int_1^0 12rsqrt(r^2+1) dr$

As the answer by VNVDVI says, the integral is $- (8 \sqrt{2} - 4) = 4 - 8 \sqrt{2}$ $-(8sqrt2-4) = 4-8sqrt2$ .

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Calc 2 Problem. Could someone help me understand the "u" notation in this integral?

2 Answers

Explanation:

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