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# Numerical integration for SQA Higher Maths 1. Calculating area under a curve

The area bounded between two curves can be found by subtracting the lower function by the upper one and integrating between the limits. # 1

Calculate the shaded area of the given graph.

The area is given by the integral ∫⁰₋₁x³ + x² + 5 − (x²/2 + 4)dx = ∫⁰₋₁x³ + x²/2 + 1dx = [x⁴/4 + x³/6 + x]⁰₋₁ = 0⁴/4 + 0³/6 + 0 − ((−1)⁴/4 + (−1)³/6 − 1) = 11/12. # 2

Find the area between y = x³ + 3 and y = x + 1 between the points x = 1 and x = 2.

The area is given by the integral ∫²₁x³ + 3 − (x + 1)dx = ∫²₁x³ − x + 4dx = [x⁴/4 − x²/2 + 4x]²₁ = 2⁴/4 − 2²/2 + 4(2) − (1⁴/4 − 1²/2 + 4(1)) = 25/4. # 3

Use definite integrals to find the area between y = 2x + 10 and y = 2x + 4 between the points x = 5 and x = 2.

The area is given by the integral ∫⁵₂2x + 10 − (2x + 4)dx = ∫⁵₂6dx = [6x]⁵₂ = 6(5) − 6(2) = 18. # 4

Evaluate the area between the curves y = −x + 9 and y = x² + 3 between the points x = 0 and x = 2.

The area is given by the integral ∫²₀−x + 9 − (x² + 3)dx = ∫²₀−x + 6 − x²dx = [−x²/2 + 6x − x³/3]²₀ = −2²/2 + 6(2) − 2³/3 − (−0²/2 + 6(0) − 0³/3) = 22/3. # 5

Find the area between y = −x and y = x² − 8 between the points x = 0 and x = 2.

The area is given by the integral ∫²₀x² − 8 − −xdx = ∫²₀x² + x − 8dx = [x³/3 + x²/2 − 8x]²₀ = 2³/3 + 2²/2 − 8(2) − (0³/3 + 0²/2 − 8(0)) = −34/3. End of page