Maths Assignment Sample
Q1:
Answer :Step 1: Solve the definite integral
∫(x2+2x+1) dx=x33+x2+x+C\int (x^2 + 2x + 1) \, dx = \frac{x^3}{3} + x^2 + x + C∫(x2+2x+1)dx=3x3+x2+x+C
Evaluate the integral from x=0x = 0x=0 to x=4x = 4x=4:
∫04(x2+2x+1) dx=[x33+x2+x]04\int_{0}^{4} (x^2 + 2x + 1) \, dx = \left[\frac{x^3}{3} + x^2 + x\right]_0^4∫04(x2+2x+1)dx=[3x3+x2+x]04
Substitute x=4x = 4x=4:
433+42+4=643+16+4=643+20=1243\frac{4^3}{3} + 4^2 + 4 = \frac{64}{3} + 16 + 4 = \frac{64}{3} + 20 = \frac{124}{3}343+42+4=364+16+4=364+20=3124
Substitute x=0x = 0x=0:
033+02+0=0\frac{0^3}{3} + 0^2 + 0 = 0303+02+0=0
Thus, the area under the curve is:
1243 units2\frac{124}{3} \, \text{units}^23124units2
Step 2: Find the average value
The average value of f(x)f(x)f(x) over [0,4][0,4][0,4] is given by:
Average value=1b−a∫abf(x) dx\text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dxAverage value=b−a1∫abf(x)dx
Substitute a=0a = 0a=0, b=4b = 4b=4, and ∫04f(x) dx=1243\int_{0}^{4} f(x) \, dx = \frac{124}{3}∫04f(x)dx=3124:
Average value=14−0⋅1243=12412=313\text{Average value} = \frac{1}{4-0} \cdot \frac{124}{3} = \frac{124}{12} = \frac{31}{3}Average value=4−01⋅3124=12124=331
Thus, the average value is:
313 units\frac{31}{3} \, \text{units}331units