Caminho de contexto simulado do Spring REST

Eu tento definir o caminho de contexto para zombarias de primavera usando o seguinte trecho de código:

private MockMvc mockMvc;

@Before
public void setUp() {
    this.mockMvc = MockMvcBuilders.webAppContextSetup(this.context)
            .apply(documentationConfiguration(this.restDocumentation))
            .alwaysDo(document("{method-name}/{step}/",
                    preprocessRequest(prettyPrint()),
                    preprocessResponse(prettyPrint())))
            .build();
}

@Test
public void index() throws Exception {
    this.mockMvc.perform(get("/").contextPath("/api").accept(MediaTypes.HAL_JSON))
            .andExpect(status().isOk())
            .andExpect(jsonPath("_links.business-cases", is(notNullValue())));
}

Mas eu recebo o seguinte erro:

java.lang.IllegalArgumentException: requestURI [/] does not start with contextPath [/api]

O que está errado? É possível especificar o contextPath em um único local no código, por exemplo. diretamente no construtor?

editar

aqui o controlador

@RestController
@RequestMapping(value = "/business-case", produces = MediaType.APPLICATION_JSON_VALUE)
public class BusinessCaseController {
    private static final Logger LOG = LoggerFactory.getLogger(BusinessCaseController.class);

    private final BusinessCaseService businessCaseService;

    @Autowired
    public BusinessCaseController(BusinessCaseService businessCaseService) {
        this.businessCaseService = businessCaseService;
    }

    @Transactional(rollbackFor = Throwable.class, readOnly = true)
    @RequestMapping(value = "/{businessCaseId}", method = RequestMethod.GET)
    public BusinessCaseDTO getBusinessCase(@PathVariable("businessCaseId") Integer businessCaseId) {
        LOG.info("GET business-case for " + businessCaseId);
        return businessCaseService.findOne(businessCaseId);
    }
}

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