Exercise · ZW-DOUBLE-INTEGRAL-OVER-GENERAL-REGION-005 · ★☆☆ Basic
The region \(D=\{0\le y\le x\le1\}\) can also be described by integrating in \(x\) first. Which iterated integral equals \(\int_0^1\!\int_0^x f\,dy\,dx\)?
Choose one option.
A \(\int_0^1\!\int_0^y f\,dx\,dy\) B \(\int_0^1\!\int_y^1 f\,dx\,dy\) C \(\int_0^1\!\int_0^1 f\,dx\,dy\) D \(\int_0^1\!\int_1^y f\,dx\,dy\) Show answer B. \(\int_0^1\!\int_y^1 f\,dx\,dy\)← Back to the question
Explanation For each \(y\) in \([0,1]\), \(x\) runs from \(y\) (the line \(y=x\)) up to \(1\). Sketching the triangle settles it.
The double integral of \(f\) with respect to the area \(A\) over a region \(D=\{a\le x\le b,\ g_1(x)\le y\le g_2(x)\}\) is an iterated integral whose inner limits depend on the outer variable.
\[\iint_D f\,dA=\int_a^b\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx\]
More practice problems: double integral over a general region ZW-DOUBLE-INTEGRAL-OVER-GENERAL-REGION-001 Compute iint_D (x+y) dA, where D=\0≤ x≤1,\ 0≤ y≤ x\.ZW-DOUBLE-INTEGRAL-OVER-GENERAL-REGION-002 Compute iint_D 2xy dA, where D=\0≤ x≤1,\ x^2≤ y≤ x\.ZW-DOUBLE-INTEGRAL-OVER-GENERAL-REGION-003 Find the area of the region between y=x^2 and y=2x, that is, iint_D dA.ZW-DOUBLE-INTEGRAL-OVER-GENERAL-REGION-004 Compute iint_D x dA, where D=\0≤ x≤2,\ 0≤ y≤ x^2\.ZW-DOUBLE-INTEGRAL-OVER-GENERAL-REGION-006 Compute ∫_0^1\!∫_y^1 sin (x^2) dx dy by reversing the order of integration.ZW-DOUBLE-INTEGRAL-OVER-GENERAL-REGION-007 Compute iint_D xy dA, where D=\0≤ y≤1,\ y≤ x≤2\.ZW-DOUBLE-INTEGRAL-OVER-GENERAL-REGION-008 Find the area of D=\1≤ x≤ e,\ 0≤ y≤ln x\.ZW-DOUBLE-INTEGRAL-OVER-GENERAL-REGION-009 Let D be the region between the graphs y=g_1(x) and y=g_2(x), with g_1≤ g_2 and a≤ x≤ b. Which is t…