Question

find the area under the curve ( y = 8x^{5} ) over the interval (0,5) give the exact value. question help: video submit question

Explanation:

Step1: Use the definite - integral formula for area

The area \(A\) under the curve \(y = f(x)\) from \(x=a\) to \(x = b\) is given by \(A=\int_{a}^{b}f(x)dx\). Here, \(f(x)=8x^{5}\), \(a = 0\), and \(b = 5\). So, \(A=\int_{0}^{5}8x^{5}dx\).

Step2: Apply the power - rule for integration

The power - rule for integration is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). For \(\int8x^{5}dx\), since \(\int k\cdot f(x)dx=k\int f(x)dx\) (\(k = 8\) is a constant), we have \(8\int x^{5}dx\). Using the power - rule, \(8\times\frac{x^{5 + 1}}{5+1}=\frac{8x^{6}}{6}=\frac{4x^{6}}{3}\).

Step3: Evaluate the definite integral

By the fundamental theorem of calculus \(\int_{a}^{b}F^{\prime}(x)dx=F(b)-F(a)\). Here, \(F(x)=\frac{4x^{6}}{3}\), so \(F(5)-F(0)=\frac{4\times5^{6}}{3}-\frac{4\times0^{6}}{3}\).
Since \(0^{6}=0\), we only need to calculate \(\frac{4\times5^{6}}{3}\). We know that \(5^{6}=15625\), then \(\frac{4\times15625}{3}=\frac{62500}{3}\).

Answer:

\(\frac{62500}{3}\)