Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — rewrite, then integrate term by term
Find ∫(6x2−x34+x)dx.
- 1∫(6x2−4x−3+x1/2)dxRewrite every term as a power of x before applying the rule.
- 26x2→36x3=2x3Add one to the index, then divide by the new index 3.
- 3−4x−3→−2−4x−2=2x−2The new index is −3+1=−2. Dividing a negative by a negative makes the term positive.
- 4x1/2→3/2x3/2=32x3/2Dividing by 23 is the same as multiplying by 32.
- 5=2x3+x22+32xx+cConvert back, and do not lose the constant of integration. Differentiate this to check it returns the original.
Example 2 — a definite integral, and the area it represents
Find the area of the region bounded by the curve y=x2+1, the x-axis, and the lines x=0 and x=3.
- 1Area=∫03(x2+1)dxThe curve is above the x-axis throughout, so the integral gives the area directly.
- 2=[3x3+x]03Integrate, then put the result in square brackets with the limits. No +c — it cancels.
- 3=(327+3)−(0+0)Top limit first, then subtract the value at the bottom limit.
- 4=9+3=12 square unitsState the units. Drag the slider on the graph above to 3.00 and you should see the same number.
Example 3 — finding a curve from its gradient
A curve has gradient dxdy=3x2−4x+1 and passes through the point (2,5). Find the equation of the curve.
- 1y=∫(3x2−4x+1)dx=x3−2x2+x+cIntegrate term by term. The constant is unknown for now.
- 2Substitute (2,5): 5=8−8+2+cPut both coordinates in — x=2 on the right, y=5 on the left.
- 35=2+c⇒c=3Solve for c. Notice it is not equal to the y-coordinate.
- 4y=x3−2x2+x+3Write the full equation out. Check by substituting x=2: 8−8+2+3=5.