Calculate the double integral. $$\iint _ { R } \left( 6 x ^ { 2 } y ^ { 3 } - 10 x ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}$$
A) 190
B) 170
C) 130
D) 150
E) 120

Question

Quizplus · Accepted Answer

To solve the double integral $\iint _ { R } \left( 6 x ^ { 2 } y ^ { 3 } - 10 x ^ { 4 } \right) d A$ over the region $R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}$, we integrate with respect to $y$ first (from 0 to 4), and then with respect to $x$ (from 0 to 1).First, integrate with respect to $y$:$$\int_{0}^{4} (6x^2y^3 - 10x^4) dy = 6x^2 \left[\frac{y^4}{4}\right]_0^4 - 10x^4 \left[y\right]_0^4$$$$= 6x^2 \left(\frac{4^4}{4}\right) - 10x^4(4)$$$$= 6x^2(64) - 40x^4$$$$= 384x^2 - 40x^4$$Next, integrate with respect to $x$:$$\int_{0}^{1} (384x^2 - 40x^4) dx = 384 \left[\frac{x^3}{3}\right]_0^1 - 40 \left[\frac{x^5}{5}\right]_0^1$$$$= 384 \left(\frac{1^3}{3}\right) - 40 \left(\frac{1^5}{5}\right)$$$$= 128 - 8$$$$= 120$$Therefore, the correct answer is $120$, which corresponds to choice E.

Calculate the Double Integral ∬R(6x2y3−10x4)dA,R={(x,y)∣0≤x≤1,0≤y≤4}\iint _ { R } \left( 6 x ^ { 2 } y ^ { 3 } - 10 x ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}∬R​(6x2y3−10x4)dA,R={(x,y)∣0≤x≤1,0≤y≤4}

Calculate the Double Integral $\iint _ { R } \left( 6 x ^ { 2 } y ^ { 3 } - 10 x ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}$