Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find the area under the graph of 
f over the interval 
[-2,4].

f(x)={[x^(2)+6,x <= 2],[5x,x > 2]:}

Find the area under the graph of f f over the interval [2,4] [-2,4] .\newlinef(x)={x2+6amp;x25xamp;xgt;2 f(x)=\left\{\begin{array}{ll} x^{2}+6 &amp; x \leq 2 \\ 5 x &amp; x&gt;2 \end{array}\right.

Full solution

Q. Find the area under the graph of f f over the interval [2,4] [-2,4] .\newlinef(x)={x2+6x25xx>2 f(x)=\left\{\begin{array}{ll} x^{2}+6 & x \leq 2 \\ 5 x & x>2 \end{array}\right.
  1. Identify Intervals and Functions: Identify the intervals and corresponding functions for the piecewise function f(x)f(x). The function f(x)f(x) is defined as x2+6x^2 + 6 for x2x \leq 2, and as 5x5x for x > 2. We need to find the area under the graph of f(x)f(x) from x=2x = -2 to x=4x = 4. This means we will have to split the integral at x=2x = 2, where the definition of the function changes.
  2. Set Up Integral for First Interval: Set up the integral for the first interval from x=2x = -2 to x=2x = 2. For x2x \leq 2, f(x)=x2+6f(x) = x^2 + 6. We will integrate this function from 2-2 to 22. The integral is 22(x2+6)dx\int_{-2}^{2} (x^2 + 6) \, dx.
  3. Calculate Integral for First Interval: Calculate the integral for the first interval. 22(x2+6)dx=[x33+6x]22\int_{-2}^{2} (x^2 + 6) \, dx = \left[\frac{x^3}{3} + 6x\right]_{-2}^{2}. Evaluating this from 2-2 to 22 gives: (233+62)((2)33+6(2))\left(\frac{2^3}{3} + 6\cdot 2\right) - \left(\frac{(-2)^3}{3} + 6\cdot (-2)\right).
  4. Perform Calculations for First Interval: Perform the calculations for the first interval.\newline(23/3+62)((2)3/3+6(2))=(8/3+12)(8/312)(2^3/3 + 6\cdot2) - ((-2)^3/3 + 6\cdot(-2)) = (8/3 + 12) - (-8/3 - 12).\newlineSimplify the expression: (8/3+12+8/3+12)=(16/3+24)=(16/3+72/3)=88/3(8/3 + 12 + 8/3 + 12) = (16/3 + 24) = (16/3 + 72/3) = 88/3.
  5. Set Up Integral for Second Interval: Set up the integral for the second interval from x=2x = 2 to x=4x = 4. For x > 2, f(x)=5xf(x) = 5x. We will integrate this function from 22 to 44. The integral is 245xdx\int_{2}^{4} 5x \, dx.
  6. Calculate Integral for Second Interval: Calculate the integral for the second interval.\newline245xdx=[5x22]24\int_{2}^{4} 5x \, dx = \left[\frac{5x^2}{2}\right]_{2}^{4}.\newlineEvaluating this from 22 to 44 gives: (5422)(5222)\left(\frac{5\cdot4^2}{2}\right) - \left(\frac{5\cdot2^2}{2}\right).
  7. Perform Calculations for Second Interval: Perform the calculations for the second interval.\newline(5×42/2)(5×22/2)=(5×16/2)(5×4/2)=(40)(10)=30(5\times4^2/2) - (5\times2^2/2) = (5\times16/2) - (5\times4/2) = (40) - (10) = 30.
  8. Add Areas for Total Area: Add the areas from both intervals to find the total area under the graph of f(x)f(x) from x=2x = -2 to x=4x = 4. Total area = Area from first interval + Area from second interval. Total area = 883+30\frac{88}{3} + 30.
  9. Convert Second Area and Add: Convert the second area to a fraction with a common denominator and add the two areas.\newlineTotal area = 883+(30×33)=883+903=1783\frac{88}{3} + \left(30 \times \frac{3}{3}\right) = \frac{88}{3} + \frac{90}{3} = \frac{178}{3}.
  10. Convert Total Area: Convert the total area to a mixed number or decimal if necessary.\newlineTotal area = 1783=5913\frac{178}{3} = 59 \frac{1}{3}.

More problems from Find derivatives of logarithmic functions