Sketch the bounded region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width.
y=x^2, y=7x^2, 3x+y=4, x>=0
Then find the area of the region
Don't need drawings or anything. Need Help please! Thanks
y=x^2, y=7x^2, 3x+y=4, x>=0
Then find the area of the region
Don't need drawings or anything. Need Help please! Thanks
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y = 7x² and y = -3x+4 intersection in QI at x = 4/7
y = x² and y = -3x+4 intersection in QI at x = 1
The bounded region has vertices (0,0), (4/7, 16/7) and (1,1).
We need to split up the integral into two pieces: one from x = 0 to 4/7, and the second piece from x = 4/7 to 1:
∫7x² - x² dx, from x = 0 to 4/7
= ∫6x² dx, from x = 0 to 4/7
= 2x³, from x = 0 to 4/7
= 128/343
+
∫(-3x+4) - x² dx, from x = 4/7 to 1
= -3x²/2 + 4x - x³/3, from x = 4/7 to 1
= 297/686
128/343 + 297/686 = 79/98
y = x² and y = -3x+4 intersection in QI at x = 1
The bounded region has vertices (0,0), (4/7, 16/7) and (1,1).
We need to split up the integral into two pieces: one from x = 0 to 4/7, and the second piece from x = 4/7 to 1:
∫7x² - x² dx, from x = 0 to 4/7
= ∫6x² dx, from x = 0 to 4/7
= 2x³, from x = 0 to 4/7
= 128/343
+
∫(-3x+4) - x² dx, from x = 4/7 to 1
= -3x²/2 + 4x - x³/3, from x = 4/7 to 1
= 297/686
128/343 + 297/686 = 79/98
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[int_0^(4/7) (7x^2 - x^2) dx] + [int_(4/7)^1 (4 - 3x - x^2) dx] = 79/98