Evaluate the definite integral \[\int_{-1}^2 (x+1)^2 \,dx .\]

Find the value of the constant \(a\) so that \[\int_{-1}^2 a(x+1)^2 \,dx = 1 .\]

Sketch graphs to illustrate how these definite integrals are related.

Probably the easiest way to evaluate this integral is to expand the bracket (later you will learn about other ways to do it). \[\int_{-1}^2 x^2+2x+1\,dx =\left[\frac{x^3}{3}+x^2+x\right]_{-1}^{2}=\left(\frac{8}{3}+4+2\right)-\left(-\frac{1}{3}+1-1\right)=9\]

If we want \[\int_{-1}^2 a(x+1)^2 \,dx = 1 \] then we need our integral to be \(9\) times smaller. We can do this by scaling (i.e. vertically stretching) our function by a factor of \(\dfrac{1}{9}\). So \(a=\dfrac{1}{9}\).

graph of the curve, the scaled curve and the area under it

We have used the fact that for a function \(f(x)\), \[a\int f(x)\,dx = \int af(x)\,dx.\] Is this always true?