Draw the region and evaluate the following integral

(a) $\int 01 \int 12 (y + 2 x) d y d x "> \int_{0}^{1} \int_{1}^{2} (y + 2 x) d y d x$

$\int 01 \int 12 (y + 2 x) d y d x ">$ (b) $\int 01 \int x 2 x x y d y d x "> \int_{0}^{1} \int_{x^{2}}^{x} x y d y d x$ (c) $\int 0 π \int y π sin x x d x d y "> \int_{0}^{π} \int_{y}^{π} \frac{sin x}{x} d x d y$

Solutions

Expert Solution

Solution : Draw the region and evaluate the following integral

(a) $\int 01 \int 12 (y + 2 x) d y d x = \int 01 (2 x + 12 y 2) | 12 d x = \int 01 (2 x + 32) d x = (x 2 + 32 x) | 01 = 52 "> \int_{0}^{1} \int_{1}^{2} (y + 2 x) d y d x = \int_{0}^{1} (2 x + \frac{1}{2} y^{2}) {∣ ∣}_{1}^{2} d x = \int_{0}^{1} (2 x + \frac{3}{2}) d x = (x^{2} + \frac{3}{2} x) {∣ ∣}_{0}^{1} = \frac{5}{2}$

Therefore, $\int 01 \int 12 (y + 2 x) d y d x = 52 ◼ "> \int_{0}^{1} \int_{1}^{2} (y + 2 x) d y d x = \frac{5}{2} ■$

(b) $\int 01 \int x 2 x x y d y d x = \int 01 12 x y 2 | x 2 x d x = \int 01 (12 x 3 - 12 x 5) d x = (18 x 4 - 112 x 6) | 01 = 124 "> \int_{0}^{1} \int_{x^{2}}^{x} x y d y d x = \int_{0}^{1} \frac{1}{2} x y^{2} {∣ ∣}_{x^{2}}^{x} d x = \int_{0}^{1} (\frac{1}{2} x^{3} - \frac{1}{2} x^{5}) d x = (\frac{1}{8} x^{4} - \frac{1}{12} x^{6}) {∣ ∣}_{0}^{1} = \frac{1}{24}$

Therefore, $\int 01 \int x 2 x x y d y d x = 124 ◼ "> \int_{0}^{1} \int_{x^{2}}^{x} x y d y d x = \frac{1}{24} ■$

(c) $\int 0 π \int y π sin x x d x d y \Rightarrow \int 0 π \int 0 x sin x x d y d x = \int 0 π sin x d x = (- cos x) | 0 π = 2 "> \int_{0}^{π} \int_{y}^{π} \frac{sin x}{x} d x d y \Rightarrow \int_{0}^{π} \int_{0}^{x} \frac{sin x}{x} d y d x = \int_{0}^{π} sin x d x = (- cos x) {∣ ∣}_{0}^{π} = 2$

Therefore, $\int 0 π \int y π sin x x d x d y = 2 ◼ "> \int_{0}^{π} \int_{y}^{π} \frac{sin x}{x} d x d y = 2 ■$

SUN BUNRA answered 1 year ago

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