Find the area under the curve y = x2 from x = 0 to x = 3. Find the area under y = 2x + 1 from x = 1 to x = 4. Find the area between y = x2 and y = x from x = 0 to x = 1.

Mathematics
Find the area under the curve y = x2 from x = 0 to x = 3. Find the area under y = 2x + 1 from x = 1 to x = 4. Find the area between y = x2 and y = x from x = 0 to x = 1.

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9

Right 🌹CONFIDENTIALITY, let's tackle these area problems.

Q1. Find the area under the curve y=x2y = x^2 from x=0x = 0 to x=3x = 3.

Step 1: Set up the definite integral for the area. A=∫03x2 dxA = \int_0^3 x^2 \, dx Step 2: Find the antiderivative of x2x^2. ∫x2 dx=x2+12+1=x33\int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} Step 3: Evaluate the antiderivative at the upper and lower limits using the Fundamental Theorem of Calculus. A=[x33]03=(333)−(033)A = \left[\frac{x^3}{3}\right]_0^3 = \left(\frac{3^3}{3}\right) - \left(\frac{0^3}{3}\right) Step 4: Calculate the numerical value. A=273−0=9A = \frac{27}{3} - 0 = 9 The final answer is 9\boxed{9}.

Q2. Find the area under y=2x+1y = 2x + 1 from x=1x = 1 to x=4x = 4.

Step 1: Set up the definite integral for the area. A=∫14(2x+1) dxA = \int_1^4 (2x + 1) \, dx Step 2: Find the antiderivative of 2x+12x + 1. ∫(2x+1) dx=2⋅x22+x=x2+x\int (2x + 1) \, dx = 2 \cdot \frac{x^2}{2} + x = x^2 + x Step 3: Evaluate the antiderivative at the upper and lower limits. A=[x2+x]14=((4)2+4)−((1)2+1)A = [x^2 + x]_1^4 = ((4)^2 + 4) - ((1)^2 + 1) Step 4: Calculate the numerical value. A=(16+4)−(1+1)=20−2=18A = (16 + 4) - (1 + 1) = 20 - 2 = 18 The final answer is 18\boxed{18}.

Q3. Find the area between y=x2y = x^2 and y=xy = x from x=0x = 0 to x=1x = 1.

Step 1: Determine which function is greater over the interval [0,1][0, 1]. For x∈(0,1)x \in (0, 1), x>x2x > x^2. For example, if x=0.5x=0.5, then x=0.5x=0.5 and x2=0.25x^2=0.25. So y=xy=x is the upper curve and y=x2y=x^2 is the lower curve. Step 2: Set up the definite integral for the area between the curves. A=∫01(x−x2) dxA = \int_0^1 (x - x^2) \, dx Step 3: Find the antiderivative of x−x2x - x^2. ∫(x−x2) dx=x22−x33\int (x - x^2) \, dx = \frac{x^2}{2} - \frac{x^3}{3} Step 4: Evaluate the antiderivative at the upper and lower limits. A=[x22−x33]01=(122−133)−(022−033)A = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \left(\frac{1^2}{2} - \frac{1^3}{3}\right) - \left(\frac{0^2}{2} - \frac{0^3}{3}\right) Step 5: Calculate the numerical value. A=(12−13)−(0−0)=36−26=16A = \left(\frac{1}{2} - \frac{1}{3}\right) - (0 - 0) = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} The final answer is 16\boxed{\frac{1}{6}}.

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Right 🌹CONFIDENTIALITY, let's tackle these area problems. Q1.

Find the area under the curve y = x2 from x = 0 to x = 3. Find the area under y = 2x + 1 from x = 1 to x = 4. Find the area between y = x2 and y = x from x = 0 to x = 1.
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI Answer
Right 🌹CONFIDENTIALITY, let's tackle these area problems. Q1. Find the area under the curve y = x^2 from x = 0 to x = 3. Step 1: Set up the definite integral for the area. A = _0^3 x^2 \, dx Step 2: Find the antiderivative of x^2. x^2 \, dx = x^2+12+1 = (x^3)/(3) Step 3: Evaluate the antiderivative at the upper and lower limits using the Fundamental Theorem of Calculus. A = [(x^3)/(3)]_0^3 = ((3^3)/(3)) - ((0^3)/(3)) Step 4: Calculate the numerical value. A = (27)/(3) - 0 = 9 The final answer is 9. Q2. Find the area under y = 2x + 1 from x = 1 to x = 4. Step 1: Set up the definite integral for the area. A = _1^4 (2x + 1) \, dx Step 2: Find the antiderivative of 2x + 1. (2x + 1) \, dx = 2 · (x^2)/(2) + x = x^2 + x Step 3: Evaluate the antiderivative at the upper and lower limits. A = [x^2 + x]_1^4 = ((4)^2 + 4) - ((1)^2 + 1) Step 4: Calculate the numerical value. A = (16 + 4) - (1 + 1) = 20 - 2 = 18 The final answer is 18. Q3. Find the area between y = x^2 and y = x from x = 0 to x = 1. Step 1: Determine which function is greater over the interval [0, 1]. For x (0, 1), x > x^2. For example, if x=0.5, then x=0.5 and x^2=0.25. So y=x is the upper curve and y=x^2 is the lower curve. Step 2: Set up the definite integral for the area between the curves. A = _0^1 (x - x^2) \, dx Step 3: Find the antiderivative of x - x^2. (x - x^2) \, dx = (x^2)/(2) - (x^3)/(3) Step 4: Evaluate the antiderivative at the upper and lower limits. A = [(x^2)/(2) - (x^3)/(3)]_0^1 = ((1^2)/(2) - (1^3)/(3)) - ((0^2)/(2) - (0^3)/(3)) Step 5: Calculate the numerical value. A = ((1)/(2) - (1)/(3)) - (0 - 0) = (3)/(6) - (2)/(6) = (1)/(6) The final answer is (1)/(6). 3 done, 2 left today. You're making progress.