How do you find the definite integral of #(3x^3 +7) dx # from #[1, 5]#?

Question

Isaiah Allen · Accepted Answer

496

#color(red)(|bar(ul(color(white)(a/a)color(black)(int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a))color(white)(a/a)|)))# Integrate each term using the #color(blue)"power rule"# #int(ax^n)dx=a/(n+1)x^(n+1)# #rArrint_1^5(3x^3+7)dx=[3/4x^4+7x]_1^5# #=(3/4(5)^4+7(5))-(3/4(1)^4+7(1))# #=1875/4+35-3/4-7=1872/4+28=496#

Emma Aldridge · Answer

undefined To find the definite integral of $3x^3 + 7$ with respect to $x$ from $1$ to $5$, you first need to find the antiderivative of the function. The antiderivative of $3x^3 + 7$ is $\frac{3}{4}x^4 + 7x$. Then, you evaluate this antiderivative at the upper limit (5) and subtract the result when evaluated at the lower limit (1).

$ \int_{1}^{5} (3x^3 + 7) \, dx = \left[ \frac{3}{4}x^4 + 7x \right]_{1}^{5} = \left( \frac{3}{4}(5)^4 + 7(5) \right) - \left( \frac{3}{4}(1)^4 + 7(1) \right) = \left( \frac{3}{4}(625) + 35 \right) - \left( \frac{3}{4}(1) + 7 \right) = \left( \frac{1875}{4} + 35 \right) - \left( \frac{3}{4} + 7 \right) = \left( \frac{1875}{4} + 35 \right) - \left( \frac{3}{4} + 7 \right) = \frac{1875}{4} + 35 - \frac{3}{4} - 7 = \frac{1875}{4} - \frac{3}{4} + 35 - 7 = \frac{1872}{4} + 28 = 468 + 28 = 496. $

So, the definite integral of $3x^3 + 7$ from $1$ to $5$ is $496$.